Question

The circle $C_1:x^2+y^2=3$, with centre at $O$, intersect 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 circle $C_2$ 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 triangle $O{R}_2{R}_3$ is $6\sqrt{2}$
• D. Area of the trianlge $P{Q}_2{Q}_3$ is $4\sqrt{2}$

Answer 1

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

$Q_2{Q}_3=12$
Area of the triangle $O{R}_2{R}_3$ is $6\sqrt{2}$
$R_2{R}_3=4\sqrt{6}$

$x^2+y^2=3,x^2=2y\Rightarrow y^2+2y3=0\Rightarrow y=1,x=\sqrt{2}$
Tangent at $(\sqrt{2},1)$ to $C_1$ is $2x+y=3$
Let centre of circle touching this tangent is $(0,y)$
Hence, $\left|\frac{\sqrt{2}\left(0\right)+y-3}{\sqrt{2+1}}\right|=2\sqrt{3}$
$|y-3|=6\Rightarrow y-3=\pm 6\Rightarrow y=-3,9$
Hence, $Q_2(0,9),Q_3(0,3),Q_2{Q}_3=12$
Now $C_2:x^2+(y9{)}^2=(2\sqrt{3}{)}^2;C_3:x^2+(y+3{)}^2=(2\sqrt{3}{)}^2$
Now foot of perpendicular from $Q_2$ and $Q_3$ on tangent
$\frac{x-0}{\sqrt{2}}=\frac{y-9}{1}=-\left(\frac{0+9-3}{3}\right)=-2\Rightarrow x=-2\sqrt{2},y=7,R_2(-2\sqrt{2},7)$
Similarly, $\frac{x-0}{\sqrt{2}}=\frac{y+3}{1}=-\left(\frac{0-3-3}{3}\right)=2\Rightarrow x=2\sqrt{2},y=-1,R_3(2\sqrt{2},-1)$
Hence, $R_2{R}_3=\sqrt{}=\sqrt{(4\sqrt{2}{)}^2+(8{)}^2}=4\sqrt{6}$
Area of $\Delta O{R}_2{R}_3=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ -2\sqrt{2}& 7& 1\\ 2\sqrt{2}& -1& 1\end{array}\right|=\left|\frac{1}{2}\right[1(2\sqrt{2}-7.2\sqrt{2})\left]\right|=6\sqrt{2}$

Area of $\Delta P{Q}_2{Q}_3=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2}& 1& 1\\ 0& 9& 1\\ 0& -3& 1\end{array}\right|=\left|\frac{1}{2}\sqrt{2}\right(9+3\left)\right|=6\sqrt{2}$