Suppose from the point $P = (m,n)$ two tangents $PQ$ and $PR$ are drawn to the points $Q$ and $R$ on the circle $x^2+y^2=a^2$. Find area of $\triangle PQR$.

The information related to this question which I know are the following:

  1. The radius of the given circle forms right-angles with the tangents at $Q$ and $R$.

  2. Area of a triangle is $\frac{1}{2}\times\text{base}\times\text{height}$.

  3. Area of the triangle joining coordinates $(a,b)$, $(c,d)$, $(e,f)$ is: $$ \frac{1}{2}\left|a(d-f)+c(f-b)+e(b-d)\right| \text{ sq.units} $$

But I cannot figure out how to use this information to solve the sum.

  • $\begingroup$ Hint. Make use of the fact that (letting $O$ be the origin), both $\triangle OPQ$ and $\triangle OPR$ are right, and therefore $\angle OPQ$ (respectively, $\angle OPR$) and $\angle POQ$ (respectively, $\angle POR$) are complementary (add up to a right angle). You may also find it easier to just treat the case $P(r, 0)$, since the more general case is covered by letting $r = \sqrt{m^2+n^2}$. $\endgroup$ – Brian Tung Jul 10 '15 at 22:21
  • $\begingroup$ Thank you.But can you please elaborate a little more about how can the 'complementary' part is to be used. $\endgroup$ – user253042 Jul 10 '15 at 22:29
  • $\begingroup$ A little busy, so for the moment, you'll have to see if you can figure it out. If I have time when I get home, I'll add some more exposition. $\endgroup$ – Brian Tung Jul 10 '15 at 22:32
  • $\begingroup$ OK Sir, i'm trying to solve it. $\endgroup$ – user253042 Jul 10 '15 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy