Question

The line joining the points $A(3,4),B(1,0)$ cuts the circle $x^2+y^2=4$ in points $P$ and $Q$. If $\frac{AP}{PQ}=\lambda$ and $\frac{AQ}{QB}=\mu$, then prove that $\lambda$ and $u$ are the roots of $3y^2+2y-21=0$.

Answer 1

Let $P$ or $Q$ divide $AB$ in the ratio $k:1$ then the co-ordinates of $P$ or $Q$ are
$(\frac{k+3}{k+1},\frac{4}{k+1})$ which lie on $x^2+y^2=4$
$\therefore (k+3{)}^2+16=4(k+4{)}^2$
or $3k^2+2k-21=0$
It gives two values of $k$, i.e. $\lambda$ and $\mu$ which correspond to the points $P$ and $Q$.