Length of tangents in circle.

Question

The chord of contact of tangents from a point $P$ to a circle passes through $Q$, if $l_1$ and $l_2$ are length of tangents from $P$ and $Q$ to the circle, then $PQ$ is equal to?

I tried using power of point but didn't helped, please give some hints.

Answer 1

One method would be as follows.

Let the circle have centre $O$ and let the 'centre' of the chord of contact be $R$.

Find $OR$ and $PR$ in terms of the radius, $r$, of the circle and $l_1$.

Then $|PQ|^2=|QR|^2+|PR|^2=r^2+ {l_2}^2-|OR|^2+|PR|^2.$

Pythagoras also gives $|PR|^2= {l_1}^2-r^2+|OR|^2$ and so $$|PQ|^2={l_1}^2+{l_2}^2.$$

Answer 2

According to the Stewart’s theorem for $\triangle QPK_2$, \begin{align} l_1^2\,n+x^2\,m&=(m+n)(l_1^2+mn) ,\\ m(x^2-l_1^2-mn-n^2)&=0 \tag{1}\label{1} . \end{align}

The power of point $Q$ provides

\begin{align} n\,(m+n)&=l_2^2 ,\\ mn+n^2&=l_2^2 \tag{2}\label{2} ,\\ \end{align}

Combination of \eqref{2} with \eqref{1} gives

\begin{align} x&=\sqrt{l_1^2+l_2^2} . \end{align}