Question

If $P$ is the point of contact of the circles
$x^{2}+y^{2}+4 x+4 y-10=0$ and
$x^{2}+y^{2}-6 x-6 y+10=0$ and $Q$ is their
external centre of similitude, then the equation of the circle with $P$ and $Q$ as the extremities of its diameter is

• A. $x^{2}+y^{2}+14 x+14 y-26=0$
• B. $x^{2}+y^{2}+5 x+5 y-8=0$
• C. $x^{2}+y^{2}-5 x-5 y+8=0$
• D. $x^{2}+y^{2}-14 x-14 y+26=0$

Answer 1

Answer: (D)

We have,
$x^{2}+y^{2}+4 x+4 y-10=0 $ and
$x^{2}+y^{2}-6 x-6 y+10=0$
Centre $c_{1}=(-2,-2) $ and $ r_{1}=3 \sqrt{2}$
$c_{2}=(3,3) $ and $ r_{2}=2 \sqrt{2}$
$P$ is the point of contact of circle

$\therefore P =\left(\frac{9 \sqrt{2}-4 \sqrt{2}}{5 \sqrt{2}}, \frac{9 \sqrt{2}-4 \sqrt{2}}{5 \sqrt{2}}\right) $
$P=(1,1)$
$Q$ is external centre of similitde,
$Q=\left(\frac{9 \sqrt{2}+4 \sqrt{2}}{\sqrt{2}}, \frac{9 \sqrt{2}+4 \sqrt{2}}{\sqrt{2}}\right)$
$Q=(13,13)$
Equation of circle $P O$ as diametre is
$(x-1)(x-13)+(y-1)(y-13)=0 $
$x^{2}+y^{2}-14 x-14 y+26=0$