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. $1$

Answer 1

The correct option is D $1$

Let the equation of two circles be

$x^2+y^2+2gx+2fy+c=0$ and

$x^2+y^2+2g^{{}^{\prime }}x+2f^{{}^{\prime }}y+c^{{}^{\prime }}=0$

The condition for these two circles to be orthogonal is

$2g{g}^{{}^{\prime }}+2f{f}^{{}^{\prime }}=c+c^{{}^{\prime }}$

Therefore,

$2(-\lambda )\times \frac{-4}{3}+2(-1)\times \left(\frac{29}{6}\right)=-7+0$

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

$⟹\frac{8\lambda }{3}=\frac{8}{3}$

$⟹\lambda =1$

Hence, the answer is option (D)