Question

If the acute angle between the circles $S \equiv x^2 + y^2 + 2kx + 4y - 3 = 0$ and $S^1 \equiv x^2 + y^2 - 4x + 2ky + 9 = 0$ is $\cos^{-1}\left(\frac{3}{8}\right)$ and the centre of $S^1 = 0$ lies in the first quadrant, then the radical axis of $S = 0$ and $S^1 = 0$ is:

• A. $x - 5y + 6 = 0$
• B. $x - 5y - 4 = 0$
• C. $5x - y - 6 = 0$
• D. $5x - y - 4 = 0$

Hint:

The radical axis of two circles is obtained by subtracting their equations. Use angle conditions to find unknown parameters and apply quadrant constraints to determine signs.

Answer 1

The Correct Option is A

Step 1: Identify centers and radii
For circle $S$: $x^2 + y^2 + 2kx + 4y - 3 = 0$
Center $C_1 = (-k, -2)$, radius $r_1 = \sqrt{k^2 + 4 + 3} = \sqrt{k^2 + 7}$
For circle $S^1$: $x^2 + y^2 - 4x + 2ky + 9 = 0$
Center $C_2 = (2, -k)$, radius $r_2 = \sqrt{4 + k^2 - 9} = \sqrt{k^2 - 5}$ Step 2: Use condition on acute angle $\theta$ between circles
\[ \cos \theta = \frac{|C_1C_2|^2 - r_1^2 - r_2^2}{-2 r_1 r_2} = \frac{3}{8}. \] Calculate $|C_1C_2|^2$: \[ |C_1C_2|^2 = (2 + k)^2 + (-k + 2)^2 = (k+2)^2 + (2 - k)^2 = 2k^2 + 8. \] Calculate $r_1^2 + r_2^2$: \[ r_1^2 + r_2^2 = (k^2 + 7) + (k^2 - 5) = 2k^2 + 2. \] Calculate numerator of cosine formula: \[ |C_1C_2|^2 - r_1^2 - r_2^2 = 6. \] Calculate denominator: \[ -2 r_1 r_2 = -2 \sqrt{k^2 + 7} \sqrt{k^2 - 5}. \] So: \[ \cos \theta = \frac{6}{-2 \sqrt{k^2 + 7} \sqrt{k^2 - 5}} = \frac{3}{8}. \] Multiply both sides by denominator: \[ 6 = -\frac{3}{8} \times 2 \sqrt{k^2 + 7} \sqrt{k^2 - 5} = -\frac{3}{4} \sqrt{k^2 + 7} \sqrt{k^2 - 5}. \] Since the square roots are positive, take absolute value: \[ \sqrt{k^2 + 7} \sqrt{k^2 - 5} = 8. \] Square both sides: \[ (k^2 + 7)(k^2 - 5) = 64, \] \[ k^4 + 2k^2 - 35 = 64, \] \[ k^4 + 2k^2 - 99 = 0. \] Let $x = k^2$: \[ x^2 + 2x - 99 = 0. \] Solve quadratic: \[ x = \frac{-2 \pm \sqrt{400}}{2} = \frac{-2 \pm 20}{2}. \] Two roots: \[ x_1 = 9, \quad x_2 = -11 \quad (\text{discard negative}). \] So $k^2 = 9$, $k = \pm 3$. Step 3: Check center of $S^1$ in first quadrant
Center of $S^1$ is $(2, -k)$. For first quadrant, $y>0$ implies $-k>0 \Rightarrow k<0$. So $k = -3$. Step 4: Find radical axis
Radical axis is $S - S^1 = 0$: \[ (x^2 + y^2 + 2kx + 4y - 3) - (x^2 + y^2 - 4x + 2ky + 9) = 0, \] \[ 2kx + 4y - 3 + 4x - 2ky - 9 = 0, \] \[ 2kx + 4x + 4y - 2ky - 12 = 0. \] Substitute $k = -3$: \[ 2(-3) x + 4x + 4y - 2(-3) y - 12 = 0, \] \[ -6x + 4x + 4y + 6y - 12 = 0, \] \[ -2x + 10 y - 12 = 0, \] \[ 2x - 10 y + 12 = 0, \] Divide by 2: \[ x - 5 y + 6 = 0. \]