Given equation of the circle is
![\[{{x}^{2}}~-\text{ }6x\text{ }+\text{ }{{y}^{2}}~+\text{ }12y\text{ }+\text{ }15\text{ }=\text{ }0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5bd7d37ad54b2a2e4e6a23b966443f68_l3.png "Rendered by QuickLaTeX.com")
The above equation can be written as
![\[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }-2\text{ }\left( 3 \right)\text{ }x\text{ }+\text{ }{{3}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\text{ }\left( 6 \right)\text{ }y\text{ }+\text{ }{{6}^{2}}~+\text{ }15\text{ }-\text{ }9\text{ }+\text{ }36\text{ }=\text{ }0 \\ {{\left( x\text{ }\text{ }-3 \right)}^{2}}~+\text{ }{{\left( y-\left( -6 \right) \right)}^{2}}~\text{ }30\text{ }=\text{ }0 \\ {{\left( x\text{ }\text{ }-3 \right)}^{2}}~+\text{ }{{\left( y-\left( -6 \right) \right)}^{2}}~=~{{\left( \surd 30 \right)}^{2}} \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-512bd07b023f993ee14b3d6224d7a362_l3.png "Rendered by QuickLaTeX.com")
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,
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }\text{ }-3 \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{ }\left( -6 \right) \right)}^{2}}~=~{{\left( \surd 60 \right)}^{2}} \\ {{x}^{2}}~\text{ }-6x\text{ }+\text{ }9\text{ }+\text{ }{{y}^{2}}~\text{ }12\text{ }y\text{ }+\text{ }36\text{ }=\text{ }60 \\ {{x}^{2}}~\text{ }-6x\text{ }+\text{ }{{y}^{2}}~+12y\text{ }+45\text{ }\text{ }60\text{ }=\text{ }0 \\ {{x}^{2}}~\text{ }-6x\text{ }+\text{ }{{y}^{2}}~+\text{ }12y\text{ }\text{ }-15\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf3a52ae6dd4637c7d07c3640759bc62_l3.png "Rendered by QuickLaTeX.com")
Hence, the required equation of the circle is x2 – 6x + y2 + 12y – 15 = 0.