The radius of a circle passing through the focus of parabola $x^{2} = 4 y$ and touching the parabola at $(6, 9)$ is

5

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

5 \sqrt{10}

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

5 \sqrt{5}

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

10

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

Open in App

Solution

The correct option is B $5 \sqrt{10}$
The equation of the parabola is $x^{2} = 4 y$
Its focus is $S (0, 1)$ .
The tangent at (6, 9) is $6 x = 2 (y + 9)$ .
$\Rightarrow 3 x = y + 9.$
$\Rightarrow 3 x - y - 9 = 0$
The equation of the circle touching at $(6, 9)$ can be written as $(x - 6)^{2} + (y - 9)^{2} + λ (3 x - y - 9) = 0$
It passes through (0, 1) so $36 + 64 + λ (- 10) = 0$
$\Rightarrow 10 λ = 100$
$\Rightarrow λ = 10$
$∴$ The equation of the required circle is
$x^{2} - 12 x + 36 + y^{2} - 18 y + 81 + 30 x - 10 y - 90 = 0$
$\Rightarrow x^{2} + y^{2} + 18 x - 28 y + 27 = 0$
$R = \sqrt{81 + 196 - 27} = 5 \sqrt{10}$

Suggest Corrections

Similar questions

x^{2} = 4 y

(6, 9)

x^{2} = 4 y

(6, 9)

x^{2} + y^{2} + m x - n y + 27 = 0

n - m - 1

x^{2} = 4 y

(6, 9)

x^{2} = 4 y

P

y^{2} = 16 x

P

Join BYJU'S Learning Program