A circle is concentric with the circle x^2 + | vedclass

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

Question

(A) The given circle is $x^2 + y^2 - 6x + 12y + 15 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $g = -3, f = 6, c = 15$.The radius $r_1

Vedclass · Accepted Answer

$x^2 + y^2 - 6x + 12y - 15 = 0$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 - 6x + 12y + 15 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $g = -3, f = 6, c = 15$.The radius $r_1$ of the given circle is $\sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 6^2 - 15} = \sqrt{9 + 36 - 15} = \sqrt{30}$.The area of the given circle is $A_1 = \pi r_1^2 = 30\pi$.The required circle is concentric,so its equation is $x^2 + y^2 - 6x + 12y + k = 0$.Its radius $r_2$ satisfies $r_2^2 = g^2 + f^2 - k = 45 - k$.The area of the required circle is $A_2 = \pi r_2^2 = \pi(45 - k)$.Given that $A_2 = 2A_1$,we have $\pi(45 - k) = 2(30\pi) = 60\pi$.$45 - k = 60 \Rightarrow k = -15$.Thus,the equation is $x^2 + y^2 - 6x + 12y - 15 = 0$.

$A$ circle is concentric with the circle $x^2 + y^2 - 6x + 12y + 15 = 0$ and has an area double that of the given circle. The equation of the circle is:

Similar Questions

If the equation $3x^2 + (3 - p)xy + qy^2 - 2px = 8pq$ represents a circle, then the area (in sq. units) of this circle is (in $\pi$)

$A$ circle touches both axes and its center lies in the fourth quadrant. If its radius is $1$,then its equation is:

The equation of the circle with centre at $(1, -2)$ and passing through the centre of the given circle $x^2 + y^2 + 2y - 3 = 0$ is

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

Find the equation of the circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module