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 the value of $\lambda$ is :

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

The equations of the circles are:

$x^2+y^2-2\lambda x-2y-7=0$

Hence, $g=-\lambda ,f=-1,c=-7$ ....$\left(1\right)$

$3x^2+3y^2-8x+29y=0$

Hence, $g^{\prime }=-\frac{4}{3},f^{\prime }=\frac{29}{6},c^{\prime }=0$ ....$\left(2\right)$

Since, they are given to be orthogonal circles, we have

$2g{g}^{\prime }+2f{f}^{\prime }=c+c^{\prime }$
$2(-\lambda )\left(\frac{-4}{3}\right)+2(-1)\left(\frac{29}{6}\right)=-7$

$\Rightarrow \frac{8\lambda }{3}-\frac{29}{3}=-7$

$\Rightarrow 8\lambda -29=-21$

$\Rightarrow 8\lambda =8$
$\Rightarrow$ $\lambda =1$