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10-1.Circle and System of Circles
normal
Tangents to a circle at points $P$ and $Q$ on the circle intersect at a point $R$. If $P Q=6$ and $P R=5$, then the radius of the circle is
A
$\frac{13}{3}$
B
$4$
C
$\frac{15}{4}$
D
$\frac{16}{5}$
(KVPY-2013)
Solution
(c)
Given,
$P R$ and $Q R$ are tangents.
$P Q=6$
$P R=5$
$P M=\frac{1}{2} P Q$
$P M=\frac{1}{2} \times 6=3$
In $\triangle P R M$,
$R M^2 =P R^2-P M^2=25-9=16$
$R M =4$
In $\triangle P R M, \quad \tan \theta=\frac{P M}{R M}=\frac{3}{4}$
In $\triangle P O R, \quad \tan \theta=\frac{O P}{P R}=\frac{r}{5}$
From Eqs.$(i)$ and $(ii)$, we get
$\frac{3}{4}=\frac{r}{5} \Rightarrow r=\frac{15}{4}$
Standard 11
Mathematics
