**Question:**

**Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area.**

**Solution:**

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.