373

You visited us 373 times! Enjoying our articles? Unlock Full Access!

Distance Formula

The shortest ...

Question

The shortest distance from the point $(2, - 7)$ to the circle $x^{2} + y^{2} - 14 x - 10 y - 151 = 0$ is

1

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

2

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

3

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

4

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

Open in App

Solution

The correct option is D $3$
$P (- 7, 2)$

$x^{2} + y^{2} - 10 x - 14 y - 151 = 0$

$C (- g, - f) = C (5, 7)$

$c = - 151$

$r = \sqrt{5^{2} + 7^{2} - (- 151)} = 15$

$P (- 7, 2), C (5, 7)$

$C P = \sqrt{(5 + 7)^{2} + (7 - 2)^{2}} = 13$

The shortest distance

$= r - C P$

$= 15 - 13 = 2$

Suggest Corrections

Similar questions

(2, - 7)

x^{2} + y^{2} - 14 x - 10 y - 151 = 0

M (- 7, 2)

x^{2} + y^{2} - 10 x - 14 y - 151 = 0

P (- 7, 2)

x^{2} + y^{2} - 10 x - 14 y - 151 = 0

M (- 1, 2)

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

(4, - 3)

x^{2} + y^{2} + 4 x - 10 y - 7 = 0

Join BYJU'S Learning Program