$2 x-3 y+1=0$ and $4 x-5 y-1=0$ are the equations of two diameters of the circle $S \equiv x^2+y^2+2 g x+2 f y-11=0$. Q and R are the points of contact of the tangents drawn from the point $P(-2,-2)$ to this circle. If C is the centre of the circle $S=0$, then the area (in square units) of the…

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$2 x-3 y+1=0$ and $4 x-5 y-1=0$ are the equations of two diameters of the circle $S \equiv x^2+y^2+2 g x+2 f y-11=0$. Q and R are the points of contact of the tangents drawn from the point $P(-2,-2)$ to this circle. If C is the centre of the circle $S=0$, then the area (in square units) of the quadrilateral $P Q C R$ is

Aicharya · Accepted Answer

Equation fo diameters of the circle are $\begin{aligned} & 2 x-3 y+1=0 ...(i)\\ & 4 x-5 y-1=0 ...(ii) \end{aligned}$
On solving equation (i) and (ii) we get centre if circle $\mathrm{S}=(-g,-f)=(3,4) \Rightarrow g=-3$ and $f=-4$

Radius of circle's, $\mathrm{CQ}=\sqrt{9+16+11}=6$ $\begin{aligned} & C P=\sqrt{(3+2)^2+(4+2)^2}=\sqrt{61} \\ & \therefore P Q=\sqrt{C P^2-C Q^2}=\sqrt{61-36}=\sqrt{25}=5 \end{aligned}$
Now, area of quadrilateral PQCR $=2 \times$ area of $\triangle \mathrm{CQP}$ $=2 \times \frac{1}{2} \times \mathrm{CQ} \times \mathrm{PQ}=6 \times 5=30$ sq. units

$2 x-3 y+1=0$ and $4 x-5 y-1=0$ are the equations of two diameters of the circle $S \equiv x^2+y^2+2 g x+2 f…

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