Question

$AB$ is any chord of the circle $x^{2}+y^{2}-6x-8y-11=0$ which subtends an angle of $\frac{\pi }{2}$ at $\left(\right.1,2\left.\right).$ If locus of mid-point of $AB$ is $x^{2}+y^{2}-2ax-2by-c=0,$ then $a+b+c$ is

• A. $8$
• B. $10$
• C. $6$
• D. $12$

Answer 1

Answer: (A)

We have,
$x^{2}+y^{2}-6x-8y-11=0$
$\Rightarrow \left(x - 3\right)^{2}+\left(y - 4\right)^{2}=36$
Centre $\equiv \left(3 , 4\right)$ , Radius $=6$

Let the mid-point of $AB$ is $P\left(h , k\right)$ .
Now,
$PB=\sqrt{\left(O B\right)^{2} - \left(O P\right)^{2}}$
$\Rightarrow PB=\sqrt{6^{2} - \left(h - 3\right)^{2} - \left(k - 4\right)^{2}}$
$\Rightarrow PB=\sqrt{36 - \left(h - 3\right)^{2} - \left(k - 4\right)^{2}}....\left(1\right)$
Locus of $P$ is a circle so it will passes through $M$ . Therefore, in $\Delta ABM$
$PM=PB=PA$ is equal to the circumcircle of the locus of $P$ and $P$ is the centre of the circle.
Then,
$PM=\sqrt{\left(h - 1\right)^{2} + \left(k - 2\right)^{2}}...\left(2\right)$
By $\left(1\right)$ and $\left(2\right)$ ,
$\sqrt{6^{2} - \left(h - 3\right)^{2} - \left(k - 4\right)^{2}}=\sqrt{\left(h - 1\right)^{2} + \left(k - 2\right)^{2}}$
$\Rightarrow h^{2}+k^{2}-4h-6k=3$
Hence, locus is
$x^{2}+y^{2}-4x-6y=3$
Comparing with $x^{2}+y^{2}-2ax-2by-c=0$ , we get
$a=2,b=3,c=3$
$\therefore a+b+c=8$