The locus of a point which moves such that the tangents from it to the the circles $x^{2} + y^{2} - 5 x - 3 = 0$ and $3 x^{2} + 3 y^{2} + 2 x + 4 y - 6 = 0$ are equal, is given by

2 x^{2} + 2 y^{2} + 7 x + 4 y - 3 = 0

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

17 x + 4 y + 3 = 0

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

4 x^{2} + 4 y^{2} - 3 x + 4 y - 9 = 0

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

13 x - 4 y + 15 = 0

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

Open in App

Solution

The correct option is C $17 x + 4 y + 3 = 0$
Given
$S_{1} : x^{2} + y^{2} - 5 x - 3 = 0$
$S_{2} : x^{2} + y^{2} + \frac{2 x}{3} + \frac{4 y}{3} - 2 = 0$
Let $p (x_{1}, y_{1})$ be any point
Length of tangent from $^{'} p^{'}$ to $S_{1}$ = length of tangent from $^{'} p^{'}$ to $^{'} h^{'}$
$= \sqrt{x_{1}^{2} + y_{1}^{2} - 5 x_{1} - 3} = \sqrt{x_{1}^{2} + y_{1}^{2} + \frac{2 x_{1}}{3} + \frac{4 y_{1}}{3} - 2}$
$= - 5 x_{1} - 3 = \frac{2 x_{1}}{3} + \frac{4 y_{1}}{3} - 2$
$= 17 x_{1} + 4 y_{1} + 3 = 0$
Hence, locus is $17 x + 4 y + 3 = 0$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 5 x - 3 = 0

3 x^{2} + 3 y^{2} + 3 x + 4 y - 6 = 0

Find the equation of pair of tangents drawn to the circle $x^{2} + y^{2} - 4 x + 4 y = 0$ from the point (2, 2)

The equation of the circle which intersects circles $x^{2} + y^{2} + x + 2 y + 3 = 0, x^{2} + y^{2} + 2 x + 4 y + 5 = 0$ and

$x^{2} + y^{2} - 7 x - 8 y - 9 = 0$ at right angle, will be

c_{1} : x^{2} + y^{2} - 2 x - 2 y + 3 = 0

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

x^{2} + y^{2} + z^{2} + 2 x - 4 y - 6 z + 5 = 0

2 x^{2} + 2 y^{2} + 2 z^{2} - 2 x + 4 y + 2 z + 3 = 0

Join BYJU'S Learning Program