If the radius of a circle x^{2}+y^{2}-4x+6y-k | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general equation of a circle is $x^{2}+y^{2}+2gx+2fy+c=0$,where the radius $r = \sqrt{g^{2}+f^{2}-c}$.Comparing the given equation $x^{2}+y^{2

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) The general equation of a circle is $x^{2}+y^{2}+2gx+2fy+c=0$,where the radius $r = \sqrt{g^{2}+f^{2}-c}$.Comparing the given equation $x^{2}+y^{2}-4x+6y-k=0$ with the general form:$2g = -4 \Rightarrow g = -2$$2f = 6 \Rightarrow f = 3$$c = -k$Given the radius $r = 5$,we have:$5 = \sqrt{(-2)^{2} + (3)^{2} - (-k)}$$5 = \sqrt{4 + 9 + k}$Squaring both sides:$25 = 13 + k$$k = 25 - 13 = 12$

If the radius of a circle $x^{2}+y^{2}-4x+6y-k=0$ is $5$,then $k=$

Similar Questions

If a circle passing through $(1, -2)$ has $x - y = 2$ and $2x + 3y = 14$ as its diameters,then the radius of the circle is

Which of the following equations represents a circle?

Find the equation of a circle with centre $(2, 2)$ and passing through the point $(4, 5)$.

The centre of the circle passing through the point $(0,1)$ and touching the curve $y=x^2$ at $(2,4)$ is

The radius of the circle $r = \sqrt{3} \sin \theta + \cos \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module