Question

If locus of Mid point of chord making right angle at $(c,0)$ inside the circle $x^2+y^2=a^{2}is{x}^2+y^2-cx=\frac{1}{m}(a^2-c^2)$. Find $m$

Answer 1

Locus of mid point of chord making right angle at $(c,0)$ inside the

Circle ----- given.

Let us consider O is center of circle and coordinate of O is (O,O)

chord is making right angle at $A(C,O)$

$\therefore \angle CAB={90}^o$

Let us consider midpoint P on chord CB

$\therefore CP=PB$

coordinate of P is (h,k)

Now, In $△APC$ & $△APB$

$PC=PB$

$AP=AP$ --------(common)

$\angle CAB={90}^o$ and AP is bisector of $\angle CAB$

$\therefore \angle CAP=\angle BAP=\frac{{90}^o}{2}={45}^o$

$\therefore △APC\cong △APB$

$\therefore AP=PC=PB$ --------(similar triangle)

Now, distance of AP will be

$AP=\sqrt{(h-c{)}^2+(k-o{)}^2}$

$=\sqrt{(h-c{)}^2+k^2}$

In $△DPC$

$a^{=}p{c}^2+o{p}^2$ ------------(a is radius given in question)

$\therefore a^2=p{c}^2\left(\sqrt{h^2+k^2}\right)$

$a^2=p{c}^2+(h^2+k^2)$

$p{c}^2=a^2-(h^2+k^2)$

$\therefore \sqrt{a^2-(h^2+k^2)=\sqrt{(h-c{)}^2}+k^2}$

Taking square on both side

$a^2-(x^2+y^2)=(x-c{)}^{+}{y}^2$

$a^2-x^2-y^2-x^2+2xc-c^2-y^2=0$

$a^2-c^2=2x^2+2y^2-2cx$

$a^2-c^2=2(x^2+y^2-cx)$

compare with eqn

$x^2+y^2-cx=\frac{1}{m}(a^2-c^2)$$

$\therefore m=2$