Question

If the line $px+qy=1$ touches the circle $x^2+y^2=r^2$, Prove that the point $(p,q)$ lies on the circle $x^2+y^2=r^{-2}$

Answer 1

We know for a line y = mx + c to be a tangent to a circle of radius "a"

C${}^2$ = a${}^2$ ( m${}^2$ + 1)

Here line is y = $\frac{-px}{q}+\frac{1}{q}$

and a=r, m = $\frac{-p}{q}$

$\Rightarrow$ $(\frac{1}{q}{)}^2=r^2(\frac{p^2}{q^2}+1)$

$\Rightarrow$ $\frac{1}{q^x}=r^2\frac{(p^2+q^2)}{q^2}$

$\Rightarrow$ $p^2+q^2=\frac{1}{r^2}$

Hence, we can say that (p, q) lie on circle $x^2+y^2=\frac{1}{r^2}$.