The radical centre of the three circles
$x^{2} + y^{2} = 9, x^{2} + y^{2} - 2 x - 2 y - 5 = 0$ and
$x^{2} + y^{2} + 4 x + 6 y - 19 = 0$ is

(1, - 1)

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

(1, 2)

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

(1, 1)

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

(- 1, - 1)

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

Open in App

Solution

The correct option is C $(1, 1)$
$s_{1} - s_{2} = 0 \Rightarrow$ radical axis
$\Rightarrow 2 x + 2 y - 4 = 0 - - (1)$
and $s_{1} - s_{3} = 0 \Rightarrow$ radical axis
$- 4 x - 6 y + 10 = 0 - - (2)$
from (1) & (2), point of intersection of radical axis.
$- 2 y + 2 = 0$
$y = 1$ and $x = 1$
So, radical centre is $(1, 1)$

Suggest Corrections

Similar questions

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

x^{2} + y^{2} - 2 x - 4 y - 1 = 0

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

x^{2} + y^{2} - 4 x - 6 y + 5 = 0, x^{2} + y^{2} - 2 x - 4 y - 1 = 0, x^{2} + y^{2} - 6 x - 2 y = 0

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

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

x^{2} + y^{2} + 2 x + 3 y + 1 = 0, x^{2} + y^{2} + x + y + 2 = 0, x^{2} + y^{2} + 3 x + 2 y + 1 = 0

x^{2} + y^{2} - 4 x - 6 y - 14 = 0

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

x^{2} + y^{2} - 10 x - 16 y + 7 = 0

Join BYJU'S Learning Program