Question

If the circles \( x^2+y^2-2\lambda x - 2y - 7 = 0 \) and \( 3(x^2+y^2) - 8x + 29y = 0 \) are orthogonal, then \( \lambda = \)

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

For two circles \(x^2+y^2+2g_1x+2f_1y+c_1=0\) and \(x^2+y^2+2g_2x+2f_2y+c_2=0\) to be orthogonal, the condition is \(2(g_1g_2 + f_1f_2) = c_1+c_2\). Ensure the equations are in the standard form (coefficients of \(x^2\) and \(y^2\) are 1) before identifying \(g, f, c\).

Answer 1

The Correct Option is D

Let the first circle be \( S_1: x^2+y^2-2\lambda x - 2y - 7 = 0 \).
Here, \( g_1 = -\lambda \), \( f_1 = -1 \), \( c_1 = -7 \).
Let the second circle be \( S_2: 3(x^2+y^2) - 8x + 29y = 0 \).
To use the orthogonality condition, the coefficients of \(x^2\) and \(y^2\) must be 1.
Divide by 3: \( S_2: x^2+y^2 - \frac{8}{3}x + \frac{29}{3}y = 0 \).
Here, \( g_2 = -\frac{4}{3} \), \( f_2 = \frac{29}{6} \), \( c_2 = 0 \).
The condition for orthogonality of two circles \( x^2+y^2+2g_1x+2f_1y+c_1=0 \) and \( x^2+y^2+2g_2x+2f_2y+c_2=0 \) is: \[ 2(g_1g_2 + f_1f_2) = c_1+c_2 \] Substitute the values: \[ 2\left( (-\lambda)\left(-\frac{4}{3}\right) + (-1)\left(\frac{29}{6}\right) \right) = -7+0 \] \[ 2\left( \frac{4\lambda}{3} - \frac{29}{6} \right) = -7 \] \[ \frac{8\lambda}{3} - \frac{29}{3} = -7 \] Multiply by 3: \[ 8\lambda - 29 = -7 \times 3 \] \[ 8\lambda - 29 = -21 \] \[ 8\lambda = -21 + 29 \] \[ 8\lambda = 8 \] \[ \lambda = 1 \] This matches option (4) if the value is 1.
The image shows the tick mark on option (4).