$AB$ 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 $AB$ $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

Let $C(3,4),D(1,2)$. Also, let $E(X,Y)$ be the midpoint of $AB$.

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Since $\triangle{CAE}$ is a right triangle with $$|AC|=6,\quad |CE|=\sqrt{(X-3)^2+(Y-4)^2}$$ we have $$|AE|^2=|AC|^2-|CE|^2=36-(X-3)^2-(Y-4)^2\tag1$$

Also, since we can see that $D$ is on the circle whose diameter is the line segment $AB$, we have $$|DE|=|AE|\tag 2$$

From $(1)(2)$, we have $$36-(X-3)^2-(Y-4)^2=(X-1)^2+(Y-2)^2,$$ i.e. $$X^2+Y^2-2\cdot 2X-2\cdot 3Y-3=0.$$

Hence, $a=2,b=3,c=3.$