Question

From the origin chords are drawn to the circle $(x-1{)}^2+y^2=1$. Find the equation of the locus of the middle points of these chords.

Answer 1

Let (h, k) be the mid point of chord OB.

$\therefore$ Point $B$ is $\Rightarrow$ $(2h,2k).$

Slope of $PC$ is : $\frac{k-0}{h-1}=\frac{k}{h-1.}$

Slope of $OB$ is : $\frac{2k}{2h}=\frac{k}{h.}$

In $△$ OPC & $△$ PCB

$=OC=CB=r(given,becauseradius)$

$CP=CP\left(Common\right)$

$OP=PB($\because$P$ is mid point).

$\therefore$ $△$ $OPC$ $\cong △PCB.$

Hence, $\angle$$OPC$ = $\angle$$CPB$

and $\angle$$OPC$ +$\angle$$CPB$ ${=180}^0$

(Sum of angles on a straight line ${=180}^0$).

$\Rightarrow$ $\angle$$OPC$ ${=90}^0.$

$\therefore$ Slope of PCX Slope of OB = -1

$\Rightarrow$ $\frac{k}{h-1}\times \frac{k}{h}=-1$

$\Rightarrow$ $k^2=-h(h-1)$

$\Rightarrow k^2=-h^2+h.$

$\Rightarrow k^2+h^2-h=0.$

$\therefore$ Equation of locus is :

$\Rightarrow$ $x^2+y^2-x=0.$