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 circles $C_2$ and $C_3$ at $R_2$ and $R_3$, respectively. Suppose $C_2$ and $C_3$ have equal radii$\left(2\sqrt{3}\right)$ 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 triangle $P{Q}_2{Q}_3$ is $4\sqrt{2}$

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 triangle $O{R}_2{R}_3$ is $6\sqrt{2}$

The point of intersection of the curves $x^2+y^2=3$ and $x^2=2y$ is $(\sqrt{2},1)$
The tangent at the point $(\sqrt{2},1)$ is given by: $\sqrt{2}x+y=3$
Let $Q_2\text{or}{Q}_3$ be $(0,k)$
Then, $\frac{|\sqrt{2}\left(0\right)+k-3|}{\sqrt{3}}=2\sqrt{3}$
$k=9\text{or}-3$
$Q_2(0,9)\text{and}{Q}_3(0,-3)$
$Q_2{Q}_3=12$
$R_2{R}_3=\sqrt{(Q_2{Q}_3{)}^2-(r_2+r_3{)}^2}\left(\text{where}{r}_2\&r_3\text{are the radius of}{Q}_2\&Q_3\text{respectively}\right)$
$R_2{R}_3=\sqrt{(12{)}^2-(4\sqrt{3}{)}^2}=4\sqrt{6}$

For triangle $O{R}_2{R}_3$, Let $h$ be the length of perpendicular from $O$ to the side $R_2{R}_3$
$h=\sqrt{3}$
Area of $O{R}_2{R}_3=\frac{1}{2}h\left(R_2{R}_3\right)=6\sqrt{2}$
Area of $P{Q}_2{Q}_3=\frac{1}{2}\times 12\times \sqrt{2}$$=6\sqrt{2}$