Let the circle touches the axes at (a, 0) and (0, a) and we get the radius to be |a| and centre of the circle as (a, a). This point lies on the line x – 2y = 3
![\[\begin{array}{*{35}{l}} a\text{ }-\text{ }2\left( a \right)\text{ }=\text{ }3 \\ -a\text{ }=\text{ }3 \\ a\text{ }=\text{ }-\text{ }3 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-73cfa34aab3d4fc35343e2c4572de1f1_l3.png "Rendered by QuickLaTeX.com")
Centre = (a, a) = (-3, -3) and radius of the circle(r) = |-3| = 3
As,we have circle with centre (-3, -3) and having radius 3.
Since, the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
substituting the values in the equation, we get
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }\left( -3 \right) \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{ }-\left( -3 \right) \right)}^{2}}~=\text{ }{{3}^{2}} \\ {{\left( x\text{ }+\text{ }3 \right)}^{2}}~+\text{ }{{\left( y\text{ }+\text{ }3 \right)}^{2}}~=\text{ }9 \\ {{x}^{2}}~+\text{ }6x\text{ }+\text{ }9\text{ }+\text{ }{{y}^{2}}~+\text{ }6y\text{ }+\text{ }9\text{ }=\text{ }9 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }6x\text{ }+\text{ }6y\text{ }+\text{ }9\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7bd1fd9cc5d4a0ca55a830dbf7061312_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 + 6x + 6y + 9 = 0.