Mathematics - Circle Question with Solution | TestHub

On solving$x^2+y^2=3$and$x^2=2y$we get point$P\left(\sqrt{2}, 1\right)$
Equation of tangent at P
$\sqrt{2} x+y=3$
Let$Q_2$be (0, k) and radius is$2\sqrt{3}$
$\therefore$$\left|\frac{\sqrt{2} \left(0\right)+k-3}{\sqrt{2+1}}\right|=2\sqrt{3}$
$\therefore$$k=9, -3$$Q_3$
$Q_2 \left(0, 9\right)$and$Q_3$ $(0,-3)$$\{\begin{array}{c}{C}_2:{\left(x-0\right)}^2+{\left(y-9\right)}^2=12\\ C_3:{\left(x-0\right)}^2+{\left(y+3\right)}^2=12\end{array}$
Hence$Q_2 Q_3=12$

${\Rightarrow a}^2+12=36$
$\Rightarrow a=2\sqrt{6}$
${\mathrm{R}}_2{\mathrm{R}}_3=2\mathrm{a}=4\sqrt{6}$
Perpendicular distance of origin O from$R_2{R}_3$is equal to distance of O from tangent$\sqrt{2}x+y=3$which is same as radius of circle$C_1= \sqrt{3}$
Hence area of$\mathrm{\Delta }O{R}_2{R}_3=\frac{1}{2}\times \left(R_2{R}_3\right) \sqrt{3}=\frac{1}{2}. 4\sqrt{6} . \sqrt{3}=6\sqrt{2}$
Perpendicular Distance of P from$Q_2{Q}_3= \sqrt{2}$
$\therefore$Area of$\mathrm{\Delta }P{Q}_2{Q}_3=\frac{1}{2}\times 12\times \sqrt{2}=6\sqrt{2}$