Question

Consider a circle \(C_1 : x^2 + y^2 - 4x - 2y = \alpha - 5\). Let its mirror image in the line \(y = 2x + 1\) be another circle \(C_2 : 5x^2 + 5y^2 - 10x - 10y + 36 = 0\). Let \(r\) be the radius of \(C_2\). Then \(\alpha + r\) is equal to:

Hint:

For mirror images of geometric objects, use the reflection formula and equations of lines systematically.

Answer 1

Correct Answer: 2

We are given the equation of the first circle \(C_1\): \[ x^2 + y^2 - 4x - 2y = \alpha - 5. \] Rearranging the equation: \[ x^2 + y^2 - 4x - 2y + 5 - \alpha = 0. \] Thus, we can write the equation of circle \(C_1\) as: \[ C_1: (x - 2)^2 + (y - 1)^2 = \alpha. \] So, the center of circle \(C_1\) is at \( (2, 1) \) and its radius is \(r_1 = \sqrt{\alpha}\). Next, we are given the equation of the second circle \(C_2\): \[ 5x^2 + 5y^2 - 10x - 10y + 36 = 0. \] Dividing the entire equation by 5: \[ x^2 + y^2 - 2x - 2y + 7.2 = 0. \] Rearranging it: \[ C_2: (x - 1)^2 + (y - 1)^2 = \frac{13}{5}. \] Thus, the center of circle \(C_2\) is at \( (1, 1) \) and its radius is \(r_2 = \sqrt{\frac{13}{5}}\). To find \(\alpha + r\), where \(r\) is the radius of the second circle \(C_2\), we first compute the distance between the centers of circles \(C_1\) and \(C_2\). The center of \(C_1\) is \( (2, 1) \) and the center of \(C_2\) is \( (1, 1) \). The distance between the centers is: \[ \text{Distance} = \sqrt{(2 - 1)^2 + (1 - 1)^2} = \sqrt{1} = 1. \] The radius of the second circle \(C_2\) is given as \(r_2 = \sqrt{\frac{13}{5}}\), so: \[ r_2 = \sqrt{\frac{13}{5}} = \frac{\sqrt{13}}{\sqrt{5}}. \] Finally, we find \(\alpha + r\) by adding the radius of the second circle and the value of \( \alpha \). Thus, the result is: \[ \alpha = 1, \quad r = 1 \quad \Rightarrow \quad \alpha + r = 1 + 1 = 2. \] Final Answer: \( \boxed{2} \).