Question

If $P$ is the point of contact of the circles
$x^2+y^2+4x+4y-10=0$ and
$x^2+y^2-6x-6y+10=0$ 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. $x^2+y^2+14x+14y-26=0$
• B. $x^2+y^2+5x+5y-8=0$
• C. $x^2+y^2-5x-5y+8=0$
• D. $x^2+y^2-14x-14y+26=0$

Answer 1

Answer: (D)

We have,
$x^2+y^2+4x+4y-10=0$ and
$x^2+y^2-6x-6y+10=0$
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 $PO$ as diametre is
$(x-1)(x-13)+(y-1)(y-13)=0$
$x^2+y^2-14x-14y+26=0$