Find the power of the point P(-1, 1) with respect to the circle
$x^{2} + y^{2} - 6 x + 4 y - 12 = 0$

Open in App

Solution

Fact: Power of a point $(x_{1}, y_{1})$ w.r.t. a circle $= 0$ is given by $\sqrt{S_{1}}$

Therefore power of $(- 1, 1)$ w.r.t. circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$
is $= \sqrt{(- 1)^{2} + 1^{2} - 6 (- 1) + 4 (1) - 12} = \sqrt{0} = 0$

Suggest Corrections

Similar questions

(1, - 2)

x^{2} + y^{2} = 1

(3, 2)

(0, 0)

x^{2} + y^{2} + 2 x + 4 y + 25 = 0

Find the equation of director circle of the circle whose diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area is 154 square units.

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

x^{2} + y^{2} + 6 x + 4 y - 12 = 0

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

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

Join BYJU'S Learning Program