The equation of the circle concentric with th | 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 get $g=-3, f=6, c=15$. The radius $r_1$ is $\sqrt{g^2+f^2-c} =

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 get $g=-3, f=6, c=15$. The radius $r_1$ 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 $A_1 = \pi r_1^2 = 30\pi$. Let the concentric circle be $x^2+y^2-6x+12y+k=0$. Its radius $r_2$ satisfies $r_2^2 = g^2+f^2-k = 45-k$. According to the problem,the area of the new circle is $A_2 = 2A_1 = 60\pi$. Thus,$\pi r_2^2 = 60\pi$,which implies $r_2^2 = 60$. Substituting $r_2^2 = 45-k$,we get $45-k = 60$,so $k = -15$. Therefore,the equation of the circle is $x^2+y^2-6x+12y-15=0$.

The equation of the circle concentric with the circle $x^2+y^2-6x+12y+15=0$ and having double its area is:

Similar Questions

Find the equation of the circle with center $(2, 1)$ and touching the $X$-axis.

Let a circle $C$ have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of $C$ on the line $x + y = 1$ is $\sqrt{14}$, then the square of the radius of $C$ is . . . . . .

If the lines $x + 2y - 5 = 0$ and $3x - y - 1 = 0$ denote two diameters of a circle of radius $5 \text{ units}$,then the equation of the circle is

$A(2,3)$ and $B(-1,1)$ are two points. If $P(x,y)$ is a variable point such that $\angle APB = 90^{\circ}$,then the locus of $P$ is:

For the circle $x^2 + y^2 + 3x + 3y = 0$,which of the following relations is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module