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  4. The circle C1: x2+y2=3, with centre at O, intersects the parabola x2=2 y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2 √3 and centre Q2 and Q3, respectively. If Q2 and Q3 lie on the y-axis, then

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Q. The circle $C_{1}: x^{2}+y^{2}=3$, with centre at $O$, intersects the parabola $x^{2}=2 y$ at the point $P$ in the first quadrant. Let the tangent to the circle $C_{1}$ at $P$ touches other two circles $C_{2}$ and $C_{3}$ at $R_{2}$ and $R_{3}$, respectively. Suppose $C_{2}$ and $C_{3}$ have equal radii $2 \sqrt{3}$ and centre $Q_{2}$ and $Q_{3}$, respectively. If $Q_{2}$ and $Q_{3}$ lie on the $y$-axis, then

JEE AdvancedJEE Advanced 2016

Solution:

Solving the equation of circle and parabola we get point $P (\sqrt{2}, 1)$
image
Now equation of tangent at $P (\sqrt{2}, 1)$ is $\sqrt{2} x + y =3$
Let the centre of the circle $C _{2}$ and $C _{3}$ are $Q _{2}\left(0, k _{1}\right)$ and $Q _{3}\left(0, k _{2}\right)$
Tangent also touches these circles, so
$\left|\frac{ k _{1}-\sqrt{3}}{\sqrt{2+1}}\right|=2 \sqrt{3}$
$\therefore k _{1}=9,-3$
So, $Q _{2}(0,9)$ and $Q _{3}(0,-3)$
$\therefore Q _{2} \,Q _{3}=12$
Let the point of contact $R _{2}$ is $\left(\alpha_{1}, \beta_{1}\right)$
tangent at $\left(\alpha_{1}, \beta_{1}\right)$ for circle $C _{2}$
$\alpha_{1} x +\beta_{1} y -9\left( y +\beta_{1}\right)=-69$
Comparing it with $\sqrt{2} x + y =3$
$\frac{\alpha_{1}}{\sqrt{2}}=\frac{\beta_{1}-9}{1}=\frac{9 \beta_{1}-69}{3}$
$\alpha_{1}=-2 \sqrt{2}, \beta_{1}=7$
$\therefore R _{2}(-2 \sqrt{2}, 7)$
Similarly let $R _{3}$ is $\left(\alpha_{2}, \beta_{2}\right)$
Equation of tangent at $\left(\alpha_{2}, \beta_{2}\right)$
$\alpha_{2} x +\beta_{2} y +3\left( y +\beta_{2}\right)=3$
Comparing with $\sqrt{2} x + y =3$
$\frac{\alpha_{2}}{\sqrt{2}}=\frac{\beta_{2}+3}{1}=\frac{3-3 \beta_{2}}{3} $
$\Rightarrow \alpha_{2}=2 \sqrt{2}, \beta_{2}=-1 $
$\therefore R _{3}(2 \sqrt{2},-1) $
$R _{2} R _{3}=\sqrt{(4 \sqrt{2})^{2}+64}=\sqrt{96}=4 \sqrt{6}$
Area of the triangle $OR _{2} R _{3}$
$=\frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ 0 & -2 \sqrt{2} & 7 \\ 0 & 2 \sqrt{2} & -1\end{vmatrix}=6 \sqrt{2}$
Area of the triangle $PQ _{2} Q _{3}$
$=\frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ \sqrt{2} & 0 & 0 \\ 1 & 9 & -3\end{vmatrix}=6 \sqrt{2}$