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Tangents are drawn from any point on the circle $x^2 + y^2 = R^2$ to the circle $x^2 + y^2 = r^2$. If the line joining the points of intersection of these tangents with the first circle also touch the second, then $R$ equals
$\sqrt{2} r$
$2r$
$\frac{{2r}}{{2 - \sqrt 3 }}$
$\frac{{4r}}{{3 - \sqrt 5 }}$
Solution
From point 'P' on the circle $x^{2}+y^{2}=R^{2},$ two tangents are drawn to the circle $x^{2}+y^{2}=r^{2}$
The tangents meet the first circle at points 'Q' and 'R'. Given that, QR is also a tangent to the circle $x^{2}+y^{2}=r^{2}$
$P S=P T=\sqrt{R^{2}-r^{2}}$
Similarly, $S Q=Q U=U R=T R=\sqrt{R^{2}-r^{2}}$
So, PQR is an equilateral triangle. $2 \theta=60^{\circ}$
$\theta=30^{\circ}$
$\ln \Delta P O S, \sin \theta=\frac{r}{R}$
$\sin 30^{\circ}=\frac{r}{R}$
$\frac{1}{2}=\frac{r}{R}$
$R=2 r$
