Tangents are drawn from any point on the circ | vedclass

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Question

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 circ

Vedclass · Accepted Answer

$2r$

Vedclass · Answer

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. $2 	heta=60^{\circ}$

$	heta=30^{\circ}$

$\ln \Delta P O S, \sin 	heta=\frac{r}{R}$

$\sin 30^{\circ}=\frac{r}{R}$

$\frac{1}{2}=\frac{r}{R}$

$R=2 r$

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

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