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


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


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:


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 $a,b,c$.


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