Question

The centre of family of circles cutting the family of circles $x^2 + y^2 + 2x \left( \lambda - \frac{3}{2} \right) + 6y \left( \lambda - \frac{2}{3} \right) - 8(\lambda + 4) = 0$ orthogonally, lies on

• A. x-y-1=0
• B. x+3y-4=0
• C. 4x+3y+7=0
• D. 3x-4y-1=0

Answer 1

Answer: (B)

x+3y-4=0

Answer 2

The given family of circles is

$x^2 + y^2 + 2x\left(\lambda - \frac{3}{2}\right) + 6y\left(\lambda - \frac{2}{3}\right) -8(\lambda+4)=0$.

Step 1: Expand and simplify:

Distribute the terms: $2x\left(\lambda - \frac{3}{2}\right) = 2\lambda x - 3x$,

$6y\left(\lambda - \frac{2}{3}\right) = 6\lambda y - 4y$,

$-8(\lambda+4) = -8\lambda -32$.

Thus, the equation becomes: $x^2+y^2 - 3x - 4y -32 + \lambda(2x+6y-8)=0$.

Step 2: Use the radical axis method:

Consider two circles from the family corresponding to different $\lambda$ values, say $\lambda_1$ and $\lambda_2$:

$S_1: \; x^2+y^2 - 3x - 4y -32 + \lambda_1(2x+6y-8)=0$,

$S_2: \; x^2+y^2 - 3x - 4y -32 + \lambda_2(2x+6y-8)=0$.

Subtract $S_2$ from $S_1$:

$(S_1 - S_2): \quad (\lambda_1 - \lambda_2)(2x+6y-8)=0$.

Since $\lambda_1 \neq \lambda_2$,

$2x+6y-8=0$.

Dividing by 2:

$x+3y-4=0$.

This line is the radical axis, and the centers of the circles cutting the given family orthogonally lie on this line.