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10-1.Circle and System of Circles
hard
Let $PQ$ and $RS$ be tangents at the extremeties of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $2r$ equals
A
$\sqrt {PQ.RS} $
B
$\frac{{PQ + RS}}{2}$
C
$\frac{{2PQ.\,\,RS}}{{PQ + RS}}$
D
$\sqrt {\frac{{P{Q^2} + R{S^2}}}{2}} $
(IIT-2001)
Solution
(a) $\tan \theta = \frac{{PQ}}{{PR}} = \frac{{PQ}}{{2r}}$
Also $\tan \left( {\frac{\pi }{2} – \theta } \right) = \frac{{RS}}{{2r}}$
$i.e.$, $\cot \theta = \frac{{RS}}{{2r}}$
$\tan \theta .\cot \theta = \frac{{PQ.RS}}{{4{r^2}}}$
==> $4{r^2} = PQ.RS$
==> $2r = \sqrt {(PQ)(RS)} $.
Standard 11
Mathematics
