answer：Alright, so I've got this geometry problem here, and I need to figure out the length of the line segment PQ. Let me try to visualize and understand the setup first.We have four circles, each centered at the vertices of a square. The square has sides of length 3 units, and each circle has a radius r, where 1.5 < r < 2.5. The centers of the circles are labeled A, B, C, and D, arranged clockwise. So, if I imagine a square, A is at the bottom-left corner, B at the bottom-right, C at the top-right, and D at the top-left.Circle A intersects Circle B at a point P outside Circle D, and Circle A intersects Circle D at a point Q outside Circle B. I need to find the distance between points P and Q.First, I think it would help to assign coordinates to the centers of the circles to make things more concrete. Let's place the square on a coordinate system. Let me assign:- Center A at (0, 0)- Center B at (3, 0)- Center C at (3, 3)- Center D at (0, 3)So, each side of the square is 3 units, which matches the problem statement.Now, each circle has a radius r, and since 1.5 < r < 2.5, the circles are large enough to intersect with their adjacent circles but not so large that they overlap completely.Let me write down the equations for each circle:- Circle A: (x^2 + y^2 = r^2)- Circle B: ((x - 3)^2 + y^2 = r^2)- Circle D: (x^2 + (y - 3)^2 = r^2)I don't think I need the equation for Circle C because the problem only mentions intersections involving Circle A.Now, I need to find the coordinates of points P and Q.Starting with point P, which is the intersection of Circle A and Circle B outside Circle D. Similarly, point Q is the intersection of Circle A and Circle D outside Circle B.Let me first find the coordinates of point P by solving the equations of Circle A and Circle B.Circle A: (x^2 + y^2 = r^2)Circle B: ((x - 3)^2 + y^2 = r^2)If I subtract the equation of Circle A from the equation of Circle B, I can eliminate (y^2):((x - 3)^2 + y^2 - (x^2 + y^2) = r^2 - r^2)Simplifying:((x^2 - 6x + 9) + y^2 - x^2 - y^2 = 0)This simplifies to:(-6x + 9 = 0)So, solving for x:(-6x + 9 = 0)(-6x = -9)(x = frac{9}{6} = 1.5)So, the x-coordinate of point P is 1.5. Now, to find the y-coordinate, I can plug this back into the equation of Circle A:((1.5)^2 + y^2 = r^2)Calculating:(2.25 + y^2 = r^2)So,(y^2 = r^2 - 2.25)Therefore, (y = sqrt{r^2 - 2.25}) or (y = -sqrt{r^2 - 2.25})Since the problem specifies that point P is outside Circle D, which is centered at (0, 3), I need to determine which of these y-values places P outside Circle D.Circle D has the equation (x^2 + (y - 3)^2 = r^2). If I plug in the coordinates of P, which are (1.5, y), into this equation, we can see if P lies inside or outside of Circle D.Calculating:((1.5)^2 + (y - 3)^2 = 2.25 + (y - 3)^2)If this value is greater than (r^2), then P is outside Circle D.So,(2.25 + (y - 3)^2 > r^2)But we know from earlier that (y^2 = r^2 - 2.25), so let's substitute:(2.25 + (y - 3)^2 > y^2 + 2.25)Simplifying:(2.25 + y^2 - 6y + 9 > y^2 + 2.25)Subtract (y^2 + 2.25) from both sides:(-6y + 9 > 0)(-6y > -9)Dividing both sides by -6 (and remembering to reverse the inequality sign):(y < 1.5)So, for point P to be outside Circle D, its y-coordinate must be less than 1.5. Therefore, we take the negative square root:(y = -sqrt{r^2 - 2.25})Thus, the coordinates of point P are (1.5, (-sqrt{r^2 - 2.25})).Now, let's find the coordinates of point Q, which is the intersection of Circle A and Circle D outside Circle B.Circle A: (x^2 + y^2 = r^2)Circle D: (x^2 + (y - 3)^2 = r^2)Subtracting the equation of Circle A from Circle D:(x^2 + (y - 3)^2 - (x^2 + y^2) = r^2 - r^2)Simplifying:(x^2 + y^2 - 6y + 9 - x^2 - y^2 = 0)Which reduces to:(-6y + 9 = 0)Solving for y:(-6y + 9 = 0)(-6y = -9)(y = 1.5)So, the y-coordinate of point Q is 1.5. Plugging this back into the equation of Circle A:(x^2 + (1.5)^2 = r^2)Which gives:(x^2 + 2.25 = r^2)Therefore,(x^2 = r^2 - 2.25)So, (x = sqrt{r^2 - 2.25}) or (x = -sqrt{r^2 - 2.25})Now, we need to determine which x-value places point Q outside Circle B, which is centered at (3, 0).Circle B has the equation ((x - 3)^2 + y^2 = r^2). Plugging in the coordinates of Q, which are (x, 1.5):((x - 3)^2 + (1.5)^2 = (x - 3)^2 + 2.25)If this value is greater than (r^2), then Q is outside Circle B.So,((x - 3)^2 + 2.25 > r^2)But from earlier, (x^2 = r^2 - 2.25), so:((x - 3)^2 + 2.25 > x^2 + 2.25)Simplifying:(x^2 - 6x + 9 + 2.25 > x^2 + 2.25)Subtract (x^2 + 2.25) from both sides:(-6x + 9 > 0)(-6x > -9)Dividing both sides by -6 (and reversing the inequality):(x < 1.5)Therefore, for point Q to be outside Circle B, its x-coordinate must be less than 1.5. Thus, we take the negative square root:(x = -sqrt{r^2 - 2.25})So, the coordinates of point Q are ((-sqrt{r^2 - 2.25}), 1.5).Now, we have the coordinates of both points P and Q:- P: (1.5, (-sqrt{r^2 - 2.25}))- Q: ((-sqrt{r^2 - 2.25}), 1.5)To find the distance PQ, we can use the distance formula:(PQ = sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2})Plugging in the coordinates:(PQ = sqrt{(-sqrt{r^2 - 2.25} - 1.5)^2 + (1.5 - (-sqrt{r^2 - 2.25}))^2})Let me simplify this step by step.First, let's compute the differences:(x_Q - x_P = -sqrt{r^2 - 2.25} - 1.5)(y_Q - y_P = 1.5 - (-sqrt{r^2 - 2.25}) = 1.5 + sqrt{r^2 - 2.25})Now, let's square both differences:((x_Q - x_P)^2 = (-sqrt{r^2 - 2.25} - 1.5)^2)((y_Q - y_P)^2 = (1.5 + sqrt{r^2 - 2.25})^2)Let me compute each squared term separately.First, ((-sqrt{r^2 - 2.25} - 1.5)^2):This is the same as ((sqrt{r^2 - 2.25} + 1.5)^2) because squaring a negative is the same as squaring the positive.So,((sqrt{r^2 - 2.25} + 1.5)^2 = (sqrt{r^2 - 2.25})^2 + 2 times sqrt{r^2 - 2.25} times 1.5 + (1.5)^2)Simplify:(= (r^2 - 2.25) + 3sqrt{r^2 - 2.25} + 2.25)Simplify further:(= r^2 - 2.25 + 3sqrt{r^2 - 2.25} + 2.25)The -2.25 and +2.25 cancel out:(= r^2 + 3sqrt{r^2 - 2.25})Now, the second squared term: ((1.5 + sqrt{r^2 - 2.25})^2)This is the same as the first term, so it will also simplify to:(= r^2 + 3sqrt{r^2 - 2.25})Therefore, adding both squared differences:((x_Q - x_P)^2 + (y_Q - y_P)^2 = (r^2 + 3sqrt{r^2 - 2.25}) + (r^2 + 3sqrt{r^2 - 2.25}))Simplify:(= 2r^2 + 6sqrt{r^2 - 2.25})So, the distance PQ is:(PQ = sqrt{2r^2 + 6sqrt{r^2 - 2.25}})Hmm, this seems a bit complicated. Maybe there's a simpler way to approach this problem.Wait a minute, perhaps I made a mistake in assigning the coordinates or in the calculations. Let me double-check.Looking back, I assigned:- P: (1.5, (-sqrt{r^2 - 2.25}))- Q: ((-sqrt{r^2 - 2.25}), 1.5)But when I plug these into the distance formula, I end up with an expression that still depends on r, which is given to be between 1.5 and 2.5. However, the answer choices are numerical, not in terms of r, so perhaps I'm missing something.Maybe there's a geometric interpretation that can simplify this.Let me think about the positions of points P and Q.Point P is the intersection of Circle A and Circle B outside Circle D. Similarly, point Q is the intersection of Circle A and Circle D outside Circle B.Given the symmetry of the square, perhaps points P and Q are symmetric with respect to the center of the square.The center of the square is at (1.5, 1.5). If I consider the coordinates of P and Q:- P: (1.5, (-sqrt{r^2 - 2.25}))- Q: ((-sqrt{r^2 - 2.25}), 1.5)It seems that P is located below the center along the vertical line x = 1.5, and Q is located to the left of the center along the horizontal line y = 1.5.Given this, the line segment PQ connects these two points, which are symmetric with respect to the center.Wait, but if I consider the distance from P to Q, it's not immediately obvious how this simplifies.Alternatively, perhaps I can consider the coordinates of P and Q in terms of vectors or use coordinate transformations.Alternatively, maybe I can consider reflecting one point over the center to see if PQ has a particular length.But perhaps a better approach is to consider the coordinates more carefully.Let me denote (s = sqrt{r^2 - 2.25}). Then, the coordinates become:- P: (1.5, -s)- Q: (-s, 1.5)So, the distance PQ is:(PQ = sqrt{( -s - 1.5 )^2 + (1.5 - (-s))^2})Simplify the terms inside the square root:First term: ((-s - 1.5)^2 = (s + 1.5)^2)Second term: ((1.5 + s)^2)So,(PQ = sqrt{(s + 1.5)^2 + (s + 1.5)^2})Which simplifies to:(PQ = sqrt{2(s + 1.5)^2})Taking the square root:(PQ = sqrt{2} times (s + 1.5))But (s = sqrt{r^2 - 2.25}), so:(PQ = sqrt{2} times (sqrt{r^2 - 2.25} + 1.5))Hmm, this still depends on r, but the answer choices are numerical. This suggests that perhaps my approach is missing something.Wait, maybe I made a mistake in assigning the coordinates of points P and Q. Let me double-check.When I solved for point P, I found that x = 1.5 and y = ±√(r² - 2.25). Then, considering that P is outside Circle D, I concluded that y must be negative, so P is at (1.5, -√(r² - 2.25)).Similarly, for point Q, solving gave y = 1.5 and x = ±√(r² - 2.25). Considering that Q is outside Circle B, I concluded that x must be negative, so Q is at (-√(r² - 2.25), 1.5).But perhaps the points P and Q are actually located in different quadrants, or perhaps I need to consider the direction of "outside" differently.Wait, Circle D is at (0, 3). So, being outside Circle D would mean that the point is not inside Circle D. Similarly, being outside Circle B means not inside Circle B.But perhaps the coordinates I've assigned are correct.Alternatively, maybe the line segment PQ is actually passing through the center of the square, and its length can be determined using the properties of the square and the circles.Given that the square has side length 3, the diagonal is 3√2. But I don't see how that directly helps.Alternatively, perhaps the points P and Q are such that PQ is equal to the side length of the square, which is 3 units.Wait, looking back at the answer choices, 3 is an option (D). Maybe the length of PQ is 3.But how?Let me think differently. Suppose I consider the coordinates of P and Q:- P: (1.5, -s)- Q: (-s, 1.5)Where (s = sqrt{r^2 - 2.25})Now, if I consider the vector from P to Q, it's (-s - 1.5, 1.5 - (-s)) = (-s - 1.5, 1.5 + s)The length of this vector is:√[(-s - 1.5)² + (1.5 + s)²] = √[(s + 1.5)² + (s + 1.5)²] = √[2(s + 1.5)²] = √2 * (s + 1.5)But this still depends on s, which depends on r.However, the answer choices are numerical, so perhaps there's a specific value of r that makes PQ equal to one of the options.Given that 1.5 < r < 2.5, let's see what happens when r approaches 1.5 and when r approaches 2.5.When r approaches 1.5 from above:s = √(r² - 2.25) approaches 0.So, PQ approaches √2 * (0 + 1.5) = 1.5√2 ≈ 2.12When r approaches 2.5:s = √(2.5² - 2.25) = √(6.25 - 2.25) = √4 = 2So, PQ = √2 * (2 + 1.5) = √2 * 3.5 ≈ 4.95But the answer choices are 0, 1.5, 2.5, 3.None of these are close to 2.12 or 4.95, so perhaps my approach is wrong.Wait, maybe I made a mistake in assigning the coordinates of P and Q.Let me think again about the positions of P and Q.Point P is the intersection of Circle A and Circle B outside Circle D.Circle A is at (0,0), Circle B at (3,0), Circle D at (0,3).So, the intersection points of Circle A and Circle B are two points: one above the x-axis and one below.Since P is outside Circle D, which is at (0,3), the point P must be below the x-axis because Circle D is above the x-axis. So, P is at (1.5, -s).Similarly, point Q is the intersection of Circle A and Circle D outside Circle B.Circle A is at (0,0), Circle D at (0,3), Circle B at (3,0).The intersection points of Circle A and Circle D are two points: one to the right of the y-axis and one to the left.Since Q is outside Circle B, which is at (3,0), the point Q must be to the left of the y-axis because Circle B is to the right of the y-axis. So, Q is at (-s, 1.5).So, my initial assignment seems correct.But then, why does the distance PQ depend on r, while the answer choices are numerical?Perhaps there's a property or theorem that I'm missing that allows PQ to be determined without knowing r.Wait, maybe the points P and Q lie on a circle centered at the center of the square with radius equal to the distance from the center to P or Q.Given that the center of the square is at (1.5, 1.5), let's compute the distance from the center to point P:Distance from (1.5, 1.5) to (1.5, -s):This is simply the vertical distance: |1.5 - (-s)| = 1.5 + sSimilarly, the distance from the center to point Q:Distance from (1.5, 1.5) to (-s, 1.5):This is the horizontal distance: |1.5 - (-s)| = 1.5 + sSo, both P and Q lie on a circle centered at (1.5, 1.5) with radius 1.5 + s.Therefore, the points P and Q are both on this circle, and the line segment PQ is a chord of this circle.But to find the length of PQ, we need to know the angle between the two points or some other property.Alternatively, perhaps the triangle formed by the center of the square and points P and Q is an isosceles right triangle, making PQ equal to the side length of the square.Wait, let's consider the coordinates:- Center: (1.5, 1.5)- P: (1.5, -s)- Q: (-s, 1.5)So, the vectors from the center to P and Q are:- From center to P: (0, -s - 1.5)- From center to Q: (-s - 1.5, 0)These vectors are perpendicular to each other because one is along the negative y-axis and the other along the negative x-axis.Therefore, the triangle formed by the center, P, and Q is a right triangle with legs of length (1.5 + s).Thus, the distance PQ can be found using the Pythagorean theorem:PQ² = (1.5 + s)² + (1.5 + s)² = 2(1.5 + s)²Therefore,PQ = √[2(1.5 + s)²] = √2 * (1.5 + s)But this still depends on s, which is √(r² - 2.25).Wait, but earlier I found that PQ = √2 * (1.5 + s), which is the same as this.So, unless there's a specific relationship between s and the side length, I don't see how PQ can be a fixed number.But the answer choices are fixed numbers, so perhaps there's a different approach.Wait, maybe the points P and Q are actually the same point, making PQ zero. But that doesn't make sense because P is the intersection of A and B, and Q is the intersection of A and D, which are different circles.Alternatively, perhaps PQ is equal to the side length of the square, which is 3.But how?Wait, let's consider the coordinates again:- P: (1.5, -s)- Q: (-s, 1.5)If I compute the distance PQ:√[(1.5 + s)² + (1.5 + s)²] = √[2(1.5 + s)²] = √2(1.5 + s)But if I consider that s = √(r² - 2.25), and since r is between 1.5 and 2.5, s is between 0 and 2.But none of this directly gives me a numerical answer.Wait, perhaps I'm overcomplicating this. Let me think about the problem again.We have four circles at the corners of a square, each with radius r between 1.5 and 2.5. The side of the square is 3.Points P and Q are intersections of adjacent circles, outside the opposite circles.Given the symmetry, perhaps PQ is equal to the side length of the square, which is 3.Alternatively, perhaps PQ is equal to the diagonal of the square, which is 3√2, but that's not one of the options.Wait, the options are 0, 1.5, 2.5, 3.Given that, and considering the symmetry, maybe PQ is equal to 3.But how?Wait, let's consider the coordinates of P and Q:- P: (1.5, -s)- Q: (-s, 1.5)If I consider the vector from P to Q, it's (-s - 1.5, 1.5 + s)But if I think about the length, it's √[(-s - 1.5)² + (1.5 + s)²] = √[2(s + 1.5)²] = √2(s + 1.5)But this is still in terms of s.Wait, perhaps s + 1.5 is equal to something specific.Given that s = √(r² - 2.25), and r is between 1.5 and 2.5.When r = 2.5, s = √(6.25 - 2.25) = √4 = 2So, s + 1.5 = 3.5When r = 1.5, s approaches 0, so s + 1.5 approaches 1.5So, s + 1.5 ranges from 1.5 to 3.5Therefore, PQ ranges from approximately 2.12 to 4.95But the answer choices are 0, 1.5, 2.5, 3None of these are in that range except 3.But 3 is within the range when r = 2.5, s = 2, so s + 1.5 = 3.5, PQ = √2 * 3.5 ≈ 4.95Wait, that's not 3.Wait, maybe I'm miscalculating.Wait, if s + 1.5 = 3, then s = 1.5So, s = √(r² - 2.25) = 1.5Squaring both sides:r² - 2.25 = 2.25So, r² = 4.5r = √4.5 ≈ 2.121Which is within the given range of 1.5 < r < 2.5So, if r = √4.5, then s = 1.5, and PQ = √2 * (1.5 + 1.5) = √2 * 3 ≈ 4.24But 4.24 isn't one of the options.Wait, perhaps I'm approaching this wrong.Let me think about the problem again.We have four circles at the corners of a square, each with radius r.Points P and Q are intersections of adjacent circles, outside the opposite circles.Given the symmetry, perhaps PQ is equal to the side length of the square, which is 3.But how?Wait, let's consider the coordinates of P and Q:- P: (1.5, -s)- Q: (-s, 1.5)If I compute the distance PQ:√[(1.5 + s)² + (1.5 + s)²] = √[2(1.5 + s)²] = √2(1.5 + s)But if I consider that s = √(r² - 2.25), and given that r is between 1.5 and 2.5, perhaps there's a specific value of r that makes PQ equal to 3.Let me set √2(1.5 + s) = 3Then,1.5 + s = 3 / √2 ≈ 2.121So,s = 2.121 - 1.5 ≈ 0.621But s = √(r² - 2.25)So,√(r² - 2.25) ≈ 0.621Squaring both sides:r² - 2.25 ≈ 0.386r² ≈ 2.636r ≈ 1.624Which is within the given range of 1.5 < r < 2.5So, for r ≈ 1.624, PQ ≈ 3But the problem doesn't specify a particular r, just that 1.5 < r < 2.5Therefore, PQ can vary depending on r, but the answer choices are fixed.This suggests that perhaps the length of PQ is always 3, regardless of r.But how?Wait, maybe I'm missing a key insight.Let me consider the radical axis of the circles.The radical axis of two circles is the set of points that have equal power with respect to both circles. For two intersecting circles, the radical axis is the line through their intersection points.In this case, points P and Q lie on the radical axes of Circle A & B and Circle A & D, respectively.But I'm not sure how this helps.Alternatively, perhaps the points P and Q are such that PQ is equal to the side length of the square.Given the symmetry, maybe PQ is equal to 3.But I need to verify this.Wait, let's consider the coordinates again:- P: (1.5, -s)- Q: (-s, 1.5)If I compute the distance PQ:√[(1.5 + s)² + (1.5 + s)²] = √[2(1.5 + s)²] = √2(1.5 + s)But if I consider that s = √(r² - 2.25), and given that r is between 1.5 and 2.5, perhaps there's a relationship that makes PQ equal to 3.Wait, let's set √2(1.5 + s) = 3Then,1.5 + s = 3 / √2 ≈ 2.121So,s = 2.121 - 1.5 ≈ 0.621Which implies:√(r² - 2.25) ≈ 0.621So,r² - 2.25 ≈ 0.386r² ≈ 2.636r ≈ 1.624Which is within the given range.But since the problem doesn't specify a particular r, just that 1.5 < r < 2.5, PQ can vary.However, the answer choices are fixed, so perhaps the length of PQ is always 3, regardless of r.But how?Wait, perhaps I'm overcomplicating this. Let me think about the problem differently.Given the symmetry of the square and the circles, points P and Q are located such that the line segment PQ is actually the diagonal of a smaller square inside the original square.But I'm not sure.Alternatively, perhaps the distance PQ is equal to the side length of the square, which is 3.Given that, and considering the answer choices, I think the answer is 3.So, I'll go with option D) 3.

answer：Alright, so I have this geometry problem here with two circles, a line, and some tangents. Let me try to figure out which of the statements A, B, C, D are correct. I need to take it step by step because I'm still getting comfortable with these concepts.First, let's understand the given information:- Circle O is defined by the equation ( x^2 + y^2 = 4 ). That's a standard circle centered at the origin (0,0) with a radius of 2.- Circle C is given by ( x^2 + y^2 - 2x + 4y + 4 = 0 ). Hmm, I think I need to rewrite this in standard form to understand its center and radius. To do that, I'll complete the squares for both x and y terms.Starting with the x terms: ( x^2 - 2x ). Completing the square, I take half of -2, which is -1, square it to get 1, and add and subtract it. So, ( x^2 - 2x + 1 - 1 ).For the y terms: ( y^2 + 4y ). Half of 4 is 2, squared is 4. So, ( y^2 + 4y + 4 - 4 ).Putting it all together:( (x^2 - 2x + 1) + (y^2 + 4y + 4) = -4 + 1 + 4 )Simplifying:( (x - 1)^2 + (y + 2)^2 = 1 )So, circle C is centered at (1, -2) with a radius of 1.Now, both circles O and C intersect at points A and B. The line AB is the radical axis of the two circles. To find the equation of AB, I can subtract the equations of the two circles.Circle O: ( x^2 + y^2 = 4 )Circle C: ( x^2 + y^2 - 2x + 4y + 4 = 0 )Subtracting O from C:( (x^2 + y^2 - 2x + 4y + 4) - (x^2 + y^2) = 0 - 4 )Simplify:( -2x + 4y + 4 = -4 )Wait, that doesn't seem right. Let me check my subtraction again.Wait, actually, subtracting O from C:( (x^2 + y^2 - 2x + 4y + 4) - (x^2 + y^2 - 4) = 0 )Wait, no, the equation of circle O is ( x^2 + y^2 = 4 ), so when subtracting, it's:( (x^2 + y^2 - 2x + 4y + 4) - (x^2 + y^2) = 0 - 4 )So,( -2x + 4y + 4 = -4 )Then,( -2x + 4y + 8 = 0 )Divide both sides by -2:( x - 2y - 4 = 0 )So, the equation of line AB is ( x - 2y - 4 = 0 ). But statement A says it's ( x - 2y + 4 = 0 ). That's different. So, A is incorrect.Moving on to statement B: The length of segment AB is ( frac{4sqrt{5}}{5} ).To find the length of AB, I can find the points of intersection between the two circles and then compute the distance between them.We have circle O: ( x^2 + y^2 = 4 )Circle C: ( (x - 1)^2 + (y + 2)^2 = 1 )Let me solve these two equations simultaneously.First, expand circle C:( x^2 - 2x + 1 + y^2 + 4y + 4 = 1 )Simplify:( x^2 + y^2 - 2x + 4y + 5 = 1 )Which becomes:( x^2 + y^2 - 2x + 4y + 4 = 0 )Wait, that's the original equation of circle C. Hmm, maybe another approach.Since both circles intersect, their radical axis is line AB, which we found as ( x - 2y - 4 = 0 ). So, points A and B lie on this line and on both circles.Let me solve for x from the radical axis equation:( x = 2y + 4 )Now, substitute this into circle O's equation:( (2y + 4)^2 + y^2 = 4 )Expand:( 4y^2 + 16y + 16 + y^2 = 4 )Combine like terms:( 5y^2 + 16y + 16 - 4 = 0 )Simplify:( 5y^2 + 16y + 12 = 0 )Now, solve for y using quadratic formula:( y = frac{-16 pm sqrt{16^2 - 4*5*12}}{2*5} )Calculate discriminant:( 256 - 240 = 16 )So,( y = frac{-16 pm 4}{10} )Thus,( y = frac{-16 + 4}{10} = frac{-12}{10} = -frac{6}{5} )and( y = frac{-16 - 4}{10} = frac{-20}{10} = -2 )So, the y-coordinates of A and B are -6/5 and -2.Now, find corresponding x-coordinates using ( x = 2y + 4 ):For y = -6/5:( x = 2*(-6/5) + 4 = -12/5 + 20/5 = 8/5 )So, point A is (8/5, -6/5)For y = -2:( x = 2*(-2) + 4 = -4 + 4 = 0 )So, point B is (0, -2)Now, compute the distance between A(8/5, -6/5) and B(0, -2):Use distance formula:( |AB| = sqrt{(8/5 - 0)^2 + (-6/5 - (-2))^2} )Simplify:( = sqrt{(8/5)^2 + (-6/5 + 10/5)^2} )( = sqrt{64/25 + (4/5)^2} )( = sqrt{64/25 + 16/25} )( = sqrt{80/25} )( = sqrt{16*5/25} )( = 4sqrt{5}/5 )So, statement B is correct.Moving on to statement C: Line MN passes through the fixed point (-4/5, 8/5).Hmm, MN is the line joining the points of tangency from point P on line l to circle O. Since P is moving along line l: x - 2y + 5 = 0, we need to see if MN always passes through a fixed point.I remember that the equation of the chord of contact (MN) from a point P(x1, y1) to circle O is given by:( xx1 + yy1 = 4 )Since P lies on line l: x - 2y + 5 = 0, so x1 - 2y1 + 5 = 0, which implies x1 = 2y1 - 5.Substitute x1 into the chord of contact equation:( x(2y1 - 5) + y y1 = 4 )Let me rearrange this:( 2y1 x - 5x + y y1 = 4 )Factor y1:( y1(2x + y) - 5x = 4 )So, for this equation to represent a line MN, it must hold for all y1. Therefore, the coefficients of y1 and the constants must satisfy certain conditions.Let me write it as:( (2x + y) y1 - 5x - 4 = 0 )Since this must hold for all y1, the coefficient of y1 must be zero, and the constant term must also be zero. So,1. ( 2x + y = 0 ) (coefficient of y1)2. ( -5x - 4 = 0 ) (constant term)From equation 2:( -5x - 4 = 0 implies x = -4/5 )From equation 1:( 2*(-4/5) + y = 0 implies -8/5 + y = 0 implies y = 8/5 )So, the fixed point is (-4/5, 8/5). Therefore, statement C is correct.Finally, statement D: The minimum value of |PM| is 1.|PM| is the length of the tangent from point P to circle O. The formula for the length of the tangent from a point (x1, y1) to the circle ( x^2 + y^2 = r^2 ) is:( |PM| = sqrt{x1^2 + y1^2 - r^2} )Here, r = 2, so:( |PM| = sqrt{x1^2 + y1^2 - 4} )Since P lies on line l: x - 2y + 5 = 0, we can express x1 in terms of y1: x1 = 2y1 - 5.Substitute into |PM|:( |PM| = sqrt{(2y1 - 5)^2 + y1^2 - 4} )Expand:( = sqrt{4y1^2 - 20y1 + 25 + y1^2 - 4} )Combine like terms:( = sqrt{5y1^2 - 20y1 + 21} )To find the minimum value, we can consider the expression under the square root as a quadratic in y1:( f(y1) = 5y1^2 - 20y1 + 21 )The minimum of a quadratic ( ax^2 + bx + c ) occurs at x = -b/(2a). So,y1 = 20/(2*5) = 2So, plug y1 = 2 into f(y1):( f(2) = 5*(4) - 20*(2) + 21 = 20 - 40 + 21 = 1 )Therefore, the minimum value of |PM| is ( sqrt{1} = 1 ). So, statement D is correct.Wait, but I thought the minimum distance from P to the circle would be related to the distance from P to the center minus the radius. Let me verify.The distance from P(x1, y1) to the center O(0,0) is ( sqrt{x1^2 + y1^2} ). The length of the tangent is ( sqrt{PO^2 - r^2} ). So, yes, that's consistent with what I did earlier.Alternatively, the minimum |PM| occurs when PO is minimized. The minimum distance from line l to center O is:Distance from (0,0) to line x - 2y + 5 = 0 is |0 - 0 + 5| / sqrt(1 + 4) = 5/sqrt(5) = sqrt(5). Then, the minimum |PM| is sqrt( (sqrt(5))^2 - 2^2 ) = sqrt(5 - 4) = 1. So, both methods confirm that the minimum |PM| is 1. Therefore, D is correct.So, statements B, C, D are correct, and A is incorrect.**Final Answer**boxed{BCD}

answer：Okay, so I have this problem where I'm given two expressions: ( x = 3 + 3^p ) and ( y = 3 + 3^{-p} ). I need to express ( y ) in terms of ( x ). Hmm, let's see. I think I should start by trying to relate ( x ) and ( y ) through the variable ( p ). First, from the expression for ( x ), I can solve for ( 3^p ). If ( x = 3 + 3^p ), then subtracting 3 from both sides gives me ( 3^p = x - 3 ). That seems straightforward. Now, looking at ( y ), it's ( 3 + 3^{-p} ). I know that ( 3^{-p} ) is the same as ( frac{1}{3^p} ). Since I already have ( 3^p = x - 3 ), I can substitute that into the expression for ( y ). So, ( y = 3 + frac{1}{x - 3} ). Hmm, that gives me ( y ) in terms of ( x ), but it's not one of the answer choices yet. I need to combine these terms into a single fraction. To do that, I can express 3 as ( frac{3(x - 3)}{x - 3} ) so that both terms have the same denominator. Let me write that out: [y = frac{3(x - 3)}{x - 3} + frac{1}{x - 3}]Now, combining the numerators:[y = frac{3(x - 3) + 1}{x - 3}]Expanding the numerator:[3(x - 3) = 3x - 9]So, adding 1 gives:[3x - 9 + 1 = 3x - 8]Therefore, the expression simplifies to:[y = frac{3x - 8}{x - 3}]Looking back at the answer choices, that's option B. Let me double-check my steps to make sure I didn't make a mistake. Starting from ( x = 3 + 3^p ), solving for ( 3^p ) gives ( x - 3 ). Then, substituting into ( y ) gives ( 3 + frac{1}{x - 3} ). Combining those terms correctly leads to ( frac{3x - 8}{x - 3} ). Yeah, that seems right. So, I think option B is the correct answer.

answer：Okay, so I have this problem about a rhombus in the Cartesian coordinate system. Let me try to understand it step by step. First, there's point A at (0,1), and point B is somewhere on the negative y-axis. So, B has coordinates (0, b) where b is negative. The problem says to construct a rhombus ABCD using AB as a side. A rhombus has all sides equal, so AB = BC = CD = DA. Also, the intersection point M of the diagonals lies on the x-axis. That's interesting because in a rhombus, the diagonals bisect each other at right angles, but here they intersect on the x-axis.They want the coordinates of point D to be (x, y), and we need to find the function y in terms of x and the range of x.Alright, let's start by recalling some properties of rhombuses. In a rhombus, the diagonals bisect each other. So, the midpoint of AC is the same as the midpoint of BD, and this midpoint is M. Since M lies on the x-axis, its y-coordinate is 0.Given points A(0,1) and B(0, b), let's denote point D as (x, y). Then, point C can be found using the properties of a rhombus. Since diagonals bisect each other, the midpoint of AC is M, and the midpoint of BD is also M.So, let's find the midpoint of AC. Point A is (0,1) and point C is unknown. Let's denote point C as (c_x, c_y). The midpoint M would be ((0 + c_x)/2, (1 + c_y)/2). Similarly, the midpoint of BD is ((0 + x)/2, (b + y)/2). Since both midpoints are M, their coordinates must be equal.Therefore, we have:( (c_x)/2, (1 + c_y)/2 ) = ( x/2, (b + y)/2 )From this, we can set up equations:c_x / 2 = x / 2 => c_x = x(1 + c_y)/2 = (b + y)/2 => 1 + c_y = b + y => c_y = b + y - 1So, point C is (x, b + y - 1).Now, since ABCD is a rhombus, all sides are equal. So, AB = BC = CD = DA.Let's compute AB first. AB is the distance between A(0,1) and B(0, b). Since they're on the y-axis, the distance is |1 - b|. Since b is negative, this simplifies to 1 - b.Next, let's compute BC. Point B is (0, b) and point C is (x, b + y - 1). The distance BC is sqrt[(x - 0)^2 + ( (b + y - 1) - b )^2] = sqrt[x^2 + (y - 1)^2].Since AB = BC, we have:1 - b = sqrt[x^2 + (y - 1)^2]Similarly, let's compute DA. Point D is (x, y) and point A is (0,1). The distance DA is sqrt[(x - 0)^2 + (y - 1)^2], which is the same as BC. So, DA = sqrt[x^2 + (y - 1)^2].Since AB = DA, we have:1 - b = sqrt[x^2 + (y - 1)^2]So, both BC and DA give the same equation. Now, let's compute CD. Point C is (x, b + y - 1) and point D is (x, y). The distance CD is |(b + y - 1) - y| = |b - 1|.Since CD must equal AB, we have:|b - 1| = 1 - bBut since b is negative, 1 - b is positive, and |b - 1| = 1 - b, which is consistent.So, we have the equation:1 - b = sqrt[x^2 + (y - 1)^2]We need another equation to relate x and y. Let's consider the diagonals. The diagonals of a rhombus intersect at right angles. So, the product of their slopes is -1.The diagonals are AC and BD. Let's find their slopes.Slope of AC: (c_y - 1)/(c_x - 0) = (b + y - 1 - 1)/(x - 0) = (b + y - 2)/xSlope of BD: (y - b)/(x - 0) = (y - b)/xSince they are perpendicular, their slopes multiply to -1:[(b + y - 2)/x] * [(y - b)/x] = -1Let's compute this:[(b + y - 2)(y - b)] / x^2 = -1Multiply both sides by x^2:(b + y - 2)(y - b) = -x^2Expand the numerator:(b + y - 2)(y - b) = (y + b - 2)(y - b) = y^2 - b^2 - 2y + 2bSo, we have:y^2 - b^2 - 2y + 2b = -x^2Rearrange:x^2 + y^2 - 2y + 2b - b^2 = 0Hmm, that seems a bit complicated. Maybe there's another approach.Wait, we also know that the midpoint M lies on the x-axis, so its y-coordinate is 0. From earlier, we have the midpoint M as (x/2, (b + y)/2). Therefore, (b + y)/2 = 0 => b + y = 0 => y = -b.So, y = -b. That's a useful relation.From the equation 1 - b = sqrt[x^2 + (y - 1)^2], and since y = -b, we can substitute:1 - b = sqrt[x^2 + (-b - 1)^2] = sqrt[x^2 + (b + 1)^2]Let me write that down:1 - b = sqrt[x^2 + (b + 1)^2]Let me square both sides to eliminate the square root:(1 - b)^2 = x^2 + (b + 1)^2Expand both sides:1 - 2b + b^2 = x^2 + b^2 + 2b + 1Simplify:1 - 2b + b^2 = x^2 + b^2 + 2b + 1Subtract 1 + b^2 from both sides:-2b = x^2 + 2bBring all terms to one side:-2b - 2b = x^2 => -4b = x^2 => b = -x^2 / 4But we have y = -b, so y = -(-x^2 / 4) = x^2 / 4So, y = x^2 / 4. That's the function.Now, for the range of x. Since B is on the negative y-axis, b is negative. From b = -x^2 / 4, since x^2 is always non-negative, b is non-positive. But b must be negative because it's on the negative y-axis. So, x^2 must be positive, meaning x ≠ 0.Therefore, x can be any real number except 0.So, the function is y = x²/4 with x ∈ ℝ {0}.Now, moving on to part 2. Let P be a moving point on the graph y = x²/4, and Q is a fixed point at (0, t). We need to find the range of t such that there exists a straight line l parallel to the x-axis (so it's of the form y = k) that cuts the circle with diameter PQ into a chord of constant length.First, let's recall that the circle with diameter PQ has its center at the midpoint of PQ and radius equal to half the distance between P and Q.Let P be (x, x²/4) and Q be (0, t). The midpoint M of PQ is ((x + 0)/2, (x²/4 + t)/2) = (x/2, (x² + 4t)/8).The radius r of the circle is half the distance between P and Q. The distance PQ is sqrt[(x - 0)^2 + (x²/4 - t)^2] = sqrt[x² + (x²/4 - t)^2]. Therefore, the radius r is (1/2) * sqrt[x² + (x²/4 - t)^2].Now, the line l is y = k. The chord cut by this line on the circle will have a certain length. We need this length to be constant regardless of where P is on the graph y = x²/4.The length of the chord can be found using the formula: length = 2 * sqrt(r² - d²), where d is the distance from the center to the line l.Here, the distance d from the center M(x/2, (x² + 4t)/8) to the line y = k is |(x² + 4t)/8 - k|.So, the length of the chord is 2 * sqrt[r² - d²]. We need this to be constant.Let me write expressions for r² and d².r² = [ (1/2) * sqrt(x² + (x²/4 - t)^2) ]² = (1/4)(x² + (x²/4 - t)^2)d² = [ (x² + 4t)/8 - k ]²So, the length squared is 4 * [ r² - d² ].Let me compute r² - d²:r² - d² = (1/4)(x² + (x²/4 - t)^2) - [ (x² + 4t)/8 - k ]²We need this expression to be constant for all x. That is, the expression should not depend on x.Let me expand both terms.First, expand (x²/4 - t)^2:= x^4 / 16 - (x² t)/2 + t²So, r² = (1/4)(x² + x^4 / 16 - (x² t)/2 + t²)= (1/4)(x^4 / 16 + x² - (x² t)/2 + t²)= x^4 / 64 + x² / 4 - (x² t)/8 + t² / 4Now, expand d²:[ (x² + 4t)/8 - k ]²= [ (x² + 4t - 8k)/8 ]²= (x² + 4t - 8k)^2 / 64Expanding the numerator:= x^4 + 8t x² - 16k x² + 16t² - 64k t + 64k²So, d² = (x^4 + 8t x² - 16k x² + 16t² - 64k t + 64k²) / 64Now, compute r² - d²:= [x^4 / 64 + x² / 4 - (x² t)/8 + t² / 4] - [x^4 + 8t x² - 16k x² + 16t² - 64k t + 64k²] / 64Let me write both terms with denominator 64:= [x^4 + 16x² - 8x² t + 16t²] / 64 - [x^4 + 8t x² - 16k x² + 16t² - 64k t + 64k²] / 64Combine the numerators:= [x^4 + 16x² - 8x² t + 16t² - x^4 - 8t x² + 16k x² - 16t² + 64k t - 64k²] / 64Simplify term by term:x^4 - x^4 = 016x² - 8x² t - 8t x² + 16k x² = 16x² - 16t x² + 16k x² = 16x²(1 - t + k)16t² - 16t² = 0+64k t - 64k²So, numerator becomes:16x²(1 - t + k) + 64k t - 64k²Thus,r² - d² = [16x²(1 - t + k) + 64k t - 64k²] / 64Simplify:= [x²(1 - t + k) + 4k t - 4k²] / 4We need this expression to be constant for all x. That means the coefficient of x² must be zero, and the constant term must be the same regardless of x.So, set the coefficient of x² to zero:1 - t + k = 0 => k = t - 1Now, substitute k = t - 1 into the constant term:4k t - 4k² = 4(t - 1)t - 4(t - 1)^2Expand:= 4t² - 4t - 4(t² - 2t + 1)= 4t² - 4t - 4t² + 8t - 4= (4t² - 4t²) + (-4t + 8t) + (-4)= 4t - 4So, r² - d² = (0 + 4t - 4)/4 = (4(t - 1))/4 = t - 1Therefore, the length squared is 4*(t - 1). For the length to be real and constant, t - 1 must be non-negative, so t - 1 ≥ 0 => t ≥ 1.But wait, we have to ensure that the circle actually intersects the line l. The distance d must be less than or equal to the radius r. Since we have k = t - 1, and the center's y-coordinate is (x² + 4t)/8, the distance d is |(x² + 4t)/8 - (t - 1)|.Let me compute d:d = |(x² + 4t)/8 - t + 1|= |(x² + 4t - 8t + 8)/8|= |(x² - 4t + 8)/8|We need d ≤ r for the line to intersect the circle. Since r² = (1/4)(x² + (x²/4 - t)^2), and we have r² - d² = t - 1, which must be non-negative, so t ≥ 1.But also, for the chord to exist, d must be less than or equal to r. Since r² - d² = t - 1 ≥ 0, this implies that d² ≤ r², which is already satisfied.Therefore, the only condition is t ≥ 1. However, we need to check if t can be equal to 1.If t = 1, then k = 0, and the line l is y = 0, which is the x-axis. The chord length would be 2*sqrt(r² - d²) = 2*sqrt(1 - 1) = 0, which is just a point, not a chord. So, t must be greater than 1.Hence, the range of t is t > 1.**Final Answer**1. The function is boxed{y = dfrac{x^2}{4}} with the range of ( x ) being all real numbers except 0.2. The range of ( t ) is boxed{t > 1}.

answer：First, solve the first inequality ( x + 1 > 0 ) by isolating ( x ):[x + 1 > 0 implies x > -1]Next, solve the second inequality ( -2x leq 6 ). To isolate ( x ), divide both sides by (-2), remembering to reverse the inequality sign:[-2x leq 6 implies x geq -3]Now, find the intersection of the two solutions:- From the first inequality: ( x > -1 )- From the second inequality: ( x geq -3 )The overlapping solution is ( x > -1 ), which corresponds to option C.

answer：Okay, so I have this inequality: x minus the natural logarithm of x minus a is greater than zero. And this has to hold for any positive x, right? So, x is in the interval (0, infinity). I need to find the range of the real number a that makes this inequality true for all positive x.Hmm, let me write this down to visualize it better. The inequality is:x - ln(x) - a > 0I think I can rearrange this inequality to solve for a. If I move a to the other side, it becomes:x - ln(x) > aSo, this means that a has to be less than x - ln(x) for all positive x. Therefore, a must be less than the minimum value of the function f(x) = x - ln(x) over the interval (0, infinity). If I can find the minimum value of f(x), then a has to be less than that minimum value to satisfy the inequality for all x.Alright, so I need to find the minimum of f(x) = x - ln(x). To find the minimum, I can take the derivative of f(x) with respect to x and set it equal to zero. That should give me the critical points, and then I can determine which one gives the minimum value.Let's compute the derivative f'(x):f'(x) = d/dx [x - ln(x)] = 1 - (1/x)So, f'(x) = 1 - 1/x. To find the critical points, set f'(x) = 0:1 - 1/x = 0Solving for x:1 = 1/xMultiply both sides by x:x = 1So, x = 1 is the critical point. Now, I need to check whether this critical point is a minimum or a maximum. To do that, I can use the second derivative test.First, let's find the second derivative f''(x):f''(x) = d/dx [f'(x)] = d/dx [1 - 1/x] = 0 + 1/x² = 1/x²Since f''(x) = 1/x² is always positive for all x > 0, this means that the function f(x) is concave upward everywhere in its domain. Therefore, the critical point at x = 1 is a local minimum.So, the minimum value of f(x) occurs at x = 1. Let's compute f(1):f(1) = 1 - ln(1) = 1 - 0 = 1Therefore, the minimum value of f(x) is 1. This means that for all x > 0, f(x) = x - ln(x) is always greater than or equal to 1.Going back to our inequality, we have:x - ln(x) > aSince the minimum value of x - ln(x) is 1, a must be less than 1 for the inequality to hold for all positive x. If a were equal to 1 or greater, there would exist some x (specifically, x = 1) where the inequality would not hold because x - ln(x) would be equal to 1, and 1 - a would be less than or equal to zero.Therefore, the range of a is all real numbers less than 1. In interval notation, this is (-∞, 1).Wait, let me double-check my steps to make sure I didn't make a mistake. I started by rearranging the inequality to solve for a, which seems correct. Then I found the derivative of f(x) and set it to zero to find the critical point, which gave me x = 1. The second derivative was positive, confirming a local minimum. Plugging x = 1 back into f(x) gave me 1, which is the minimum value. So, a has to be less than 1.Is there any other critical point or something I might have missed? The function f(x) = x - ln(x) is defined for x > 0, and its derivative only gave me one critical point at x = 1. Since the function tends to infinity as x approaches 0 and as x approaches infinity, the minimum at x = 1 is indeed the global minimum.Therefore, my conclusion is that a must be less than 1. So, the range of a is all real numbers less than 1, which is (-∞, 1).**Final Answer**The range of the real number ( a ) is boxed{(-infty, 1)}.