Find the equation

Equation of the circle,

x² - 6x + y² + 12y + 15 = 0

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 = units square

Area of outer circle = units square {Given}

So,

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.