Question

$O$ is a fixed point and $P$ any point on a fixed circle; on $OP$ is taken a point $O$ such that $OQ$ is in a constant ratio to $OP$; prove that the locus of $Q$ is a circle.

Answer 1

Let the equation of circle be $x^2+y^2=r^2$

let point Q be represented by $(h,k)$, P by $(x_2,y_2)$ and O by $(x_1,y_1)$ and the ratio $QP:OQ=n$ $\because$ OP and OQ are in fixed ratio $\Rightarrow$ QP and OP have a constant ratio too

That gives

$h=\frac{n{x}_1+x_2}{n+1};k=\frac{n{y}_1+y_2}{n+1}$

$\Rightarrow (x_2,y_2)=\left(\right(n+1)h-n{x}_1,(n+1)k-n{y}_1)$

$(x_2,y_2)$ lie on the circle $x^2+y^2=r^2$

$\therefore x_2^2+y_2^2=r^2$

$\Rightarrow \left(\right(n+1)h-n{x}_1{)}^2+((n+1)k-n{y}_1{)}^2=r^2$

$\Rightarrow (n+1{)}^2(h^2+k^2)+n^2(x_1^2+y_1^2)-2(n+1)n{x}_{1}h-2(n+1)n{y}_{1}k-r^2=0$

Replacing $(h,k)$ with $(x,y)$

$x^2+y^2-\frac{2n{x}_1}{(n+1)}x-\frac{2n{y}_1}{n+1}y+\frac{n^2(x_1^2+y_1^2)-r^2}{(n+1{)}^2}=0$

This is an equation of circle,

Hence proved, locus of Q is a circle