The equation of a circle concentric with the circle x2+y2-6 x+12 y+15=0 and having area that is twice the area of the given circle is

Question

The equation of a circle concentric with the circle x2+y2-6 x+12 y+15=0 and having area that is twice the area of the given circle is (A) x2+y2-6 x+12 y-15=0 (B) x2+y2-6 x+12 y-30=0 (C) x2+y2-6 x+12 y-60=0 (D) x2+y2-6 x+12 y+15=0

Tardigrade Admin · Accepted Answer

Given, x2+y2-6 x+12 y+15=0 Centre (3,-6) r=√9+36-15=√30 Area =π(√30)2=30 π Required equation of circle has twice the area of circle ∴ A'=60 π π R2=60 π, R2=60 ∴ Equation of circle concentric with given circle is (x-3)2+(y+6)2=60 x2+y2-6 x+12 y-15=0

Q. The equation of a circle concentric with the circle $x^{2} + y^{2} - 6 x + 12 y + 15 = 0$ and having area that is twice the area of the given circle is

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

$x^{2} + y^{2} - 6 x + 12 y - 30 = 0$

$x^{2} + y^{2} - 6 x + 12 y - 60 = 0$

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

Q. The equation of a circle concentric with the circle x2+y2−6x+12y+15=0 and having area that is twice the area of the given circle is

x2+y2−6x+12y−15=0

x2+y2−6x+12y−30=0

x2+y2−6x+12y−60=0

x2+y2−6x+12y+15=0

Given, x2+y2−6x+12y+15=0 Centre (3,−6) r=9+36−15​=30​ Area =π(30​)2=30π Required equation of circle has twice the area of circle ∴A′=60π πR2=60π,R2=60 ∴ Equation of circle concentric with given circle is (x−3)2+(y+6)2=60 x2+y2−6x+12y−15=0

Q. The equation of a circle concentric with the circle $x^{2} + y^{2} - 6 x + 12 y + 15 = 0$ and having area that is twice the area of the given circle is

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

$x^{2} + y^{2} - 6 x + 12 y - 30 = 0$

$x^{2} + y^{2} - 6 x + 12 y - 60 = 0$

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