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 $(1,2).$ 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,
$PB=\sqrt{{\left(OB\right)}^2-{\left(OP\right)}^2}$
$\Rightarrow PB=\sqrt{6^2-{\left(h-3\right)}^2-{\left(k-4\right)}^2}$
$\Rightarrow PB=\sqrt{36-{\left(h-3\right)}^2-{\left(k-4\right)}^2}....\left(1\right)$
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
$a=2,b=3,c=3$
$\therefore a+b+c=8$