Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area.
Given equation of the circle is
x2 – 6x + y2 + 12y + 15 = 0
The above equation can be written as
x2 – 2 (3) x + 32 + y2 + 2 (6) y + 62 + 15 – 9 + 36 = 0
(x – 3)2 + (y-(-6))2 – 30 = 0
(x – 3)2 + (y-(-6))2 = (√30)2
Since, the equation of a circle having centre (h, k), having radius as r units, is
(x – h)2 + (y – k)2 = r2
Centre = (3,-6)
Area of inner circle = πr2 = 22/7 × 30 = 30 π units square
Area of outer circle = 2 × 30 π = 60 π units square
So, πr2 = 60 π
r2 = 60
Equation of outer circle is,
(x – 3)2 + (y – (-6))2 = (√60)2
x2 – 6x + 9 + y2 – 12 y + 36 = 60
x2 – 6x + y2 +12y +45 – 60 = 0
x2 – 6x + y2 + 12y – 15 = 0
Hence, the required equation of the circle is x2 – 6x + y2 + 12y – 15 = 0.