Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{P}$ and $\mathrm{Q}$ be two external points of the circle $S \equiv x^2+y^2-a^2=0$. Let the chord of contact of the point $\mathrm{P}$ with respect to the circle $S=0$ pass through Q. If $l_1$ and $l_2$ are the lengths of the tangents drawn from $\mathrm{P}$ and $\mathrm{Q}$ to the circle $S=0$, then $\mathrm{PQ}=$
Options:
Solution:
2821 Upvotes
Verified Answer
The correct answer is:
$\sqrt{l_1^2+l_2^2}$
Given circle $S \equiv x^2+y^2=a^2$ with two external points $\mathrm{P}$ and Q.
Let point $\mathrm{P}$ be $(\mathrm{h}, \mathrm{k})$ and point $\mathrm{Q}$ be $(\mathrm{p}, \mathrm{q})$.
Equation of chord of contacts of tangent from $\mathrm{P}$ is $\mathrm{xh}+$ $\mathrm{yk}=\mathrm{a}^2$.
It also passes through $\mathrm{Q}(\mathrm{p}, \mathrm{q})$.
$$
\begin{aligned}
& P Q=\sqrt{(h-p)^2+(k-q)^2} \\
& =\sqrt{h^2+p^2-2 h p+k^2+q^2-2 k q} \\
& =\sqrt{\left(h^2+k^2\right)+\left(p^2+q^2\right)-2(h p+k q)} \\
& =\sqrt{\left(h^2+k^2\right)+\left(p^2+q^2\right)-2 a^2} \\
& =\sqrt{\left(h^2+k^2-a^2\right)+\left(p^2+q^2-a^2\right)} \\
& =\sqrt{l_1^2+l_2^2}
\end{aligned}
$$
So, option (c) is correct.
Let point $\mathrm{P}$ be $(\mathrm{h}, \mathrm{k})$ and point $\mathrm{Q}$ be $(\mathrm{p}, \mathrm{q})$.
Equation of chord of contacts of tangent from $\mathrm{P}$ is $\mathrm{xh}+$ $\mathrm{yk}=\mathrm{a}^2$.
It also passes through $\mathrm{Q}(\mathrm{p}, \mathrm{q})$.
$$
\begin{aligned}
& P Q=\sqrt{(h-p)^2+(k-q)^2} \\
& =\sqrt{h^2+p^2-2 h p+k^2+q^2-2 k q} \\
& =\sqrt{\left(h^2+k^2\right)+\left(p^2+q^2\right)-2(h p+k q)} \\
& =\sqrt{\left(h^2+k^2\right)+\left(p^2+q^2\right)-2 a^2} \\
& =\sqrt{\left(h^2+k^2-a^2\right)+\left(p^2+q^2-a^2\right)} \\
& =\sqrt{l_1^2+l_2^2}
\end{aligned}
$$
So, option (c) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.