From a point P on the circle x^2+y^2-4x-6y+9= | vedclass

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(C) The equations of the two concentric circles are $x^2+y^2-4x-6y+9=0$ and $x^2+y^2-4x-6y+12=0$.For the first circle,the centre is $C(2, 3)$ and the

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$\frac{\sqrt{3}}{4}$

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(C) The equations of the two concentric circles are $x^2+y^2-4x-6y+9=0$ and $x^2+y^2-4x-6y+12=0$.For the first circle,the centre is $C(2, 3)$ and the radius is $R = \sqrt{2^2+3^2-9} = \sqrt{4+9-9} = 2$.For the second circle,the centre is $C(2, 3)$ and the radius is $r = \sqrt{2^2+3^2-12} = \sqrt{4+9-12} = 1$.Point $P$ lies on the outer circle,so $PC = R = 2$.In the right-angled triangle $	riangle PQC$ (where $\angle PQC = 90^\circ$ as $PQ$ is a tangent),we have $\cos 	heta = \frac{QC}{PC} = \frac{r}{R} = \frac{1}{2}$.Thus,$	heta = \frac{\pi}{3}$.The area of $	riangle CQR$ is given by $\frac{1}{2} 	imes (CQ) 	imes (CR) 	imes \sin(2	heta)$.Since $CQ = CR = r = 1$,the area is $\frac{1}{2} 	imes 1^2 	imes \sin(2 	imes \frac{\pi}{3}) = \frac{1}{2} 	imes \sin(\frac{2\pi}{3}) = \frac{1}{2} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$ sq. units.

From a point $P$ on the circle $x^2+y^2-4x-6y+9=0$,a pair of tangents $PQ$ and $PR$ are drawn touching the circle $x^2+y^2-4x-6y+12=0$ at $Q$ and $R$. If $C$ is the centre of the concentric circles,then the area of the $\triangle CQR$ (in sq. units) is

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