The circle $C_{1} : x^{2} + y^{2} = 3$ having centre at origin,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

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

area of the $Δ O R_{2} R_{3} i s 6 \sqrt{2}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

area of the $Δ P Q_{2} Q_{3} i s 4 \sqrt{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
area of the $Δ O R_{2} R_{3} i s 6 \sqrt{2}$

Given $C_{1} : x^{2} + y^{2} = 3$ intersects the parabola $x^{2} = 2 y$

On solving $x^{2} + y^{2} = 3$ and $x^{2} = 2 y$ , we get
$y^{2} + 2 y = 3$
$\Rightarrow y^{2} + 2 y - 3 = 0$
$\Rightarrow (y + 3) (y - 1) = 0$
$∴ y = 1, - 3$ [neglecting $y = - 3, a s - \sqrt{3} \leq y \leq \sqrt{3}]$
$∴ y = 1 \Rightarrow x = \pm \sqrt{2}$
$\Rightarrow P (\sqrt{2}, 1) ϵ I q u a d r a n t$
Equation of tangent at $P (\sqrt{2}, 1)$ to $C_{1} : x^{2} + y^{2} = 3$ is
$\sqrt{2} x + 1. y = 3 \dots \dots (i)$
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 $2 \sqrt{3}$ .
$∴ \frac{| 0 + k - 3 |}{\sqrt{2 + 1}} = 2 \sqrt{3} (D i s t a n c e o f c e n t r e Q_{2} f r o m t a n g e n t l i n e) \Rightarrow | k - 3 | = 6 \Rightarrow k = 9, - 3 ∴ Q_{2} (0, 9) a n d Q_{3} (0, - 3)$
Hence, $Q_{2} Q_{3} = 12$
$∴$ Option (a) is correct.
Also $R_{2} R_{3}$ is common internal tangent to $C_{2}$ and $C_{3}$
And $r_{2} = r_{3} = 2 \sqrt{3}$
$∴ R_{2} R_{3} = \sqrt{d^{2} - (r_{1} + r_{2})^{2}} = \sqrt{12^{2} - (4 \sqrt{3})^{2}} = \sqrt{(144 - 48)} = \sqrt{96} = 4 \sqrt{6}$

$∴$ Option (b) is correct
$∵$ Length of perpendicular from O(0, 0) to $R_{2} R_{3}$ is equal to radius of $C_{1} = \sqrt{3}$ .
$∴ A r e a o f Δ O R_{2} R_{3} = \frac{1}{2} \times R_{2} R_{3} \times \sqrt{3} = \frac{1}{2} \times 4 \sqrt{6} \times \sqrt{3} = 6 \sqrt{2}$
$∴$ Option (c) is correct.
Also are $Δ P Q_{2} Q_{3} = \frac{1}{2} Q_{2} Q_{3} \times \sqrt{2} = \frac{\sqrt{2}}{2} \times 12 = 6 \sqrt{2}$
$∴$ Option (d) is incorrect

Suggest Corrections

Similar questions

The circle $C_{1} : x^{2} + y^{2} = 3$ having centre at origin,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

Q. The circle

C_{1} : x^{2} + y^{2} = 3

, with centre at

O

, intersect the parabola

x^{2} = 2 y

at the point

P

in the first quadrant. Let the tangent to the circle

C_{1}

P

touches other two circles

C_{2}