$AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$ $AB$ is any chord of the circle $x^2+y^2-6x-8y-11=0,$which subtend $90^\circ$ at $(1,2)$.If locus of mid-point of $AB$ is circle $x^2+y^2-2ax-2by-c=0$.Find $a,b,c$.

The point $(1,2)$ is inside the circle $x^2+y^2-6x-8y-11=0$.I let the points $A(x_1,y_1)$ and $B(x_2,y_2)$ are the end points of the chord $AB$.As $AB$ subtend $90^\circ$ at $(1,2)$
So $\frac{y_1-2}{x_1-1}\times \frac{y_2-2}{x_2-1}=-1$
But i do not know how to find the locus of mid point of chord $AB$ $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

Use polar coordinate. 
Let $(x_1,y_1)=(3+6\cos\theta_1,4+6\sin\theta_1), (x_2,y_2)=(3+6\cos\theta_2, 4+6\cos \theta_2)$. 
Then by your equation, we have
$$(2+6\cos \theta_1)(2+6\cos \theta_2)+(2+6\sin\theta_1)(2+6\sin\theta_2)=0$$
Using sum and difference formula, this gives us
$$18\cos(\theta_2-\theta_1)=-4-3(\cos\theta_1+\cos\theta_2)-3(\sin\theta_1+\sin\theta_2)$$
Now find the midpoint $x=3+3\cos\theta_1+3\cos\theta_2, y=4+3\sin\theta_1+3\sin\theta_2$. Compute:
$$x^2+y^2=\dots=43+18(\cos\theta_1+\cos\theta_2)+12(\sin\theta_1+\sin\theta_2)+18\cos(\theta_2-\theta_1)$$
With the above derived formula, we get
$$x^2+y^2=39+15(\cos\theta_1+\cos\theta_2)+9(\sin\theta_1+\sin\theta_2)$$
Now this is equal to $2ax+2by+c=2a(3+3(\cos\theta_1+\cos\theta_2))+2b(4+3(\sin\theta_1+\sin\theta_2))+c$
So you can find $a,b,c$.
