Question

The value of \( k \), if the circles \( 2x^2 + 2y^2 - 4x + 6y = 3 \) and \( x^2 + y^2 + kx + y = 0 \) cut orthogonally is:

• A. \( 2 \)
• B. \( 3 \)
• C. \( 4 \)
• D. \( 5 \)
• E. \( 1 \)

Hint:

For orthogonal circles, always convert equations to the form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), then use: \[ 2(g_1 g_2 + f_1 f_2) = c_1 + c_2 \]

Answer 1

The Correct Option is A

Concept: Two circles cut orthogonally if: \[ 2(g_1 g_2 + f_1 f_2) = c_1 + c_2 \] for circles in the form: \[ x^2 + y^2 + 2g x + 2f y + c = 0 \]

Step 1: Convert both equations into standard form. First circle: \[ 2x^2 + 2y^2 - 4x + 6y = 3 \] Divide by 2: \[ x^2 + y^2 - 2x + 3y - \frac{3}{2} = 0 \] So: \[ g_1 = -1, f_1 = \frac{3}{2}, c_1 = -\frac{3}{2} \] Second circle: \[ x^2 + y^2 + kx + y = 0 \] So: \[ g_2 = \frac{k}{2}, f_2 = \frac{1}{2}, c_2 = 0 \]

Step 2: Apply orthogonality condition. \[ 2(g_1 g_2 + f_1 f_2) = c_1 + c_2 \] Substitute values: \[ 2\left[(-1)\left(\frac{k}{2}\right) + \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)\right] = -\frac{3}{2} \] \[ 2\left[-\frac{k}{2} + \frac{3}{4}\right] = -\frac{3}{2} \] \[ - k + \frac{3}{2} = -\frac{3}{2} \]

Step 3: Solve for \(k\). \[ - k = -\frac{3}{2} - \frac{3}{2} = -3 \] \[ k = 3 \]