The circle , with centre at O, intersects the parabola at the point P in the first quadrant. Let the tangent to the circle , at P touches other two circles and at and , respectively. Suppose and have equal radii and centres and , respectively. If

Question

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 centres $Q_{2}$ and $Q_{3}$ , respectively. If $Q_{2}$ and $Q_{3}$ lie on the $y$ -axis, then [2016]

Smart Achievers · Accepted Answer

(1, 2, 3)

Given circle, $C_{1} : x^{2} + y^{2} = 3 ...(i)$

and parabola: $x^{2} = 2 y ...(ii)$

Intersection point of (i) and (ii) in first quadrant

$C_{3}$

$∴$ $x = \sqrt{2} \Rightarrow P (\sqrt{2}, 1)$

Equation of tangent to circle $C_{1}$ at $P$ is $\sqrt{2} x + y - 3 = 0$

Let centre of circle $C_{2}$ be $(0, k)$ ; $r = 2 \sqrt{3}$

$∴$ $| \frac{k - 3}{\sqrt{3}} |$ $= 2 \sqrt{3} \Rightarrow k = 9 or - 3$

$∴$ $Q_{2} (0, 9), Q_{3} (0, - 3)$

$(1) Q_{2} Q_{3} = 12$

$(2) R_{2} R_{3} = length of transverse common tangent$ $= \sqrt{{(Q_{2} Q_{3})}^{2} - {(r_{1} + r_{2})}^{2}} = \sqrt{{(12)}^{2} - {(4 \sqrt{3})}^{2}} = 4 \sqrt{6}$

$(3) Area (∆ O R_{2} R_{3}) = \frac{1}{2} \times R_{2} R_{3} \times length of ⊥ from origin to tangent$
$= \frac{1}{2} \times 4 \sqrt{6} \times \sqrt{3} = 6 \sqrt{2}$

$(4) ar (∆ P Q_{2} Q_{3}) = \frac{1}{2} \times Q_{2} Q_{3} ⊥ distance of P from y-axis$
$= \frac{1}{2} \times 12 \times \sqrt{2} = 6 \sqrt{2}$

Solution

1 $Q_{2} Q_{3} = 12$

2 $R_{2} R_{3} = 4 \sqrt{6}$

3 area of the triangle $O R_{2} R_{3}$ is $6 \sqrt{2}$

4 area of the triangle $P Q_{2} Q_{3}$ is $4 \sqrt{2}$

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Solution

1 Q2Q3=12

2 R2R3=46

3 area of the triangle OR2R3 is 62

4 area of the triangle PQ2Q3 is 42

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1 $Q_{2} Q_{3} = 12$

2 $R_{2} R_{3} = 4 \sqrt{6}$

3 area of the triangle $O R_{2} R_{3}$ is $6 \sqrt{2}$

4 area of the triangle $P Q_{2} Q_{3}$ is $4 \sqrt{2}$