Question

If the angle between the circles \( x^2+y^2-2x+ky+1=0 \) and \( x^2+y^2-kx-2y+1=0 \) is \( \cos^{-1}\left(\frac{1}{4}\right) \) and \( k<0 \) then the point which lies on the radical axis of the given circles is

• A. \( (1,-3) \)
• B. \( (-1,3) \)
• C. \( (-1,-3) \)
• D. \( (1,3) \)

Hint:

The radical axis is always perpendicular to the line joining the centers.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

Find \( k \) using the angle of intersection formula. Then determine the radical axis equation \( S_1 - S_2 = 0 \) and verify which point lies on it. 
Step 2: Detailed Explanation:

Circle 1:
\( g_1 = -1, f_1 = k/2, c_1 = 1 \). Radius \( r_1 = \sqrt{1 + k^2/4 - 1} = |k|/2 \). 

Circle 2:
\( g_2 = -k/2, f_2 = -1, c_2 = 1 \). Radius \( r_2 = \sqrt{k^2/4 + 1 - 1} = |k|/2 \). Angle \( \theta \) is given by \( \cos \theta = 1/4 \). Formula: \[ \cos \theta = \frac{c_1 + c_2 - 2g_1g_2 - 2f_1f_2}{2r_1r_2} \] \[ \frac{1}{4} = \frac{1 + 1 - 2(-1)(-k/2) - 2(k/2)(-1)}{2(k/2)(k/2)} \] \[ \frac{1}{4} = \frac{2 - k + k}{k^2/2} = \frac{2}{k^2/2} = \frac{4}{k^2} \] \[ k^2 = 16 \implies k = -4 \text{ (since } k < 0 \text{)} \] 

Radical Axis \( S_1 - S_2 = 0 \):
\[ (-2x + ky) - (-kx - 2y) = 0 \] \[ (k - 2)x - (k + 2)y \text{ is incorrect subtraction direction, let's simplify:} \] \[ S_1 - S_2: (-2 - (-k))x + (k - (-2))y = 0 \] \[ (k - 2)x + (k + 2)y = 0 \text{ is WRONG.} \] Correct: \[ -2x + ky + 1 - (-kx - 2y + 1) = 0 \] \[ (k-2)x + (k+2)y = 0 \] Substitute \( k = -4 \): \[ (-4-2)x + (-4+2)y = 0 \implies -6x - 2y = 0 \implies 3x + y = 0 \] Check options: (A) \( (1, -3) \implies 3(1) - 3 = 0 \). (Satisfies) (B) \( (-1, 3) \implies -3 + 3 = 0 \). (Satisfies) The provided answer is (A). Note that both satisfy the equation derived. Usually, there's a unique answer, so let's re-verify signs. \( S_1: -2x -4y \). \( S_2: 4x -2y \). \( S_1 - S_2 = (-2-4)x + (-4 - (-2))y = -6x - 2y = 0 \implies 3x + y = 0 \). Both options A and B satisfy. Following the exam key, the answer is A. 
Step 3: Final Answer:

The point is \( (1, -3) \).