Question

The circle $C_1:x^2+y^2=3$ having centre at origin,O intersects the parabola $x^2=2y$ 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 centres $Q_2$ and $Q_3$, respectively. If $Q_2$ and $Q_3$ lie on the Y-axis then

• A. $Q_2{Q}_3=12$
• B. $R_2{R}_3=4\sqrt{6}$
• C. area of the $\Delta O{R}_2{R}_{3}is6\sqrt{2}$
• D. area of the $\Delta P{Q}_2{Q}_{3}is4\sqrt{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$Q_2{Q}_3=12$

B

$R_2{R}_3=4\sqrt{6}$

C

area of the $\Delta O{R}_2{R}_{3}is6\sqrt{2}$

Given $C_1:x^2+y^2=3$ intersects the parabola $x^2=2y$


On solving $x^2+y^2=3$ and $x^2=2y$, we get
$y^2+2y=3$
$\Rightarrow y^2+2y-3=0$
$\Rightarrow (y+3)(y-1)=0$
$\therefore y=1,-3$ [neglecting $y=-3,as-\sqrt{3}\le y\le \sqrt{3}]$
$\therefore y=1\Rightarrow x=\pm \sqrt{2}$
$\Rightarrow P(\sqrt{2},1)ϵIquadrant$
Equation of tangent at $P(\sqrt{2},1)$ to $C_1:x^2+y^2=3$ is
$\sqrt{2}x+1.y=3\dots \dots \left(i\right)$
Now, let the centres of $C_2$ and $C_3$ be $Q_2$ and $Q_3$, and tangent at P touches $C_2$ and $C_3$ at $R_2$ and $R_3$ shown as below



Let $Q_2$ be (0, k) and radius is $2\sqrt{3}$.
$\therefore \frac{|0+k-3|}{\sqrt{2+1}}=2\sqrt{3}\left(Distanceofcentre{Q}_{2}fromtangentline\right)\Rightarrow |k-3|=6\Rightarrow k=9,-3\therefore Q_2(0,9)and{Q}_3(0,-3)$
Hence, $Q_2{Q}_3=12$
$\therefore$ Option (a) is correct.
Also $R_2{R}_3$ is common internal tangent to $C_2$ and $C_3$
And $r_2=r_3=2\sqrt{3}$
$\therefore R_2{R}_3=\sqrt{d^2-(r_1+r_2{)}^2}=\sqrt{{12}^2-(4\sqrt{3}{)}^2}=\sqrt{(144-48)}=\sqrt{96}=4\sqrt{6}$


$\therefore$ Option (b) is correct
$\because$ Length of perpendicular from O(0, 0) to $R_2{R}_3$ is equal to radius of $C_1=\sqrt{3}$.
$\therefore Areaof\Delta O{R}_2{R}_3=\frac{1}{2}\times R_2{R}_3\times \sqrt{3}=\frac{1}{2}\times 4\sqrt{6}\times \sqrt{3}=6\sqrt{2}$
$\therefore$ Option (c) is correct.
Also are $\Delta P{Q}_2{Q}_3=\frac{1}{2}{Q}_2{Q}_3\times \sqrt{2}=\frac{\sqrt{2}}{2}\times 12=6\sqrt{2}$
$\therefore$ Option (d) is incorrect