The diameters (2, -3) and (-2, 4).
By using the formula,
Centre = (-a, -b)
![\[\begin{array}{*{35}{l}} =\text{ }\left[ \left( 2-2 \right)/2,\text{ }\left( -3+4 \right)/2 \right] \\ =\text{ }\left( 0,\text{ }{\scriptscriptstyle 1\!/\!{ }_2} \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3ebab340f06bc77762939f80fc67800c_l3.png "Rendered by QuickLaTeX.com")
By using the distance formula,
So, r =
![\[\begin{array}{*{35}{l}} =~\surd [{{\left( 2-0 \right)}^{2}}~+\text{ }{{\left( -3-{\scriptscriptstyle 1\!/\!{ }_2} \right)}^{2}}] \\ =~\surd [{{\left( 2 \right)}^{2}}~+\text{ }{{\left( -7/2 \right)}^{2}}] \\ =~\surd \left[ 4\text{ }+\text{ }49/4 \right] \\ =~\surd \left[ 65/4 \right] \\ =\text{ }\left[ \surd 65 \right]/2 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d7e9daec3a145a956d75256b5d5128b1_l3.png "Rendered by QuickLaTeX.com")
We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
Now by substituting the values in the above equation, we get
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }0 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }{\scriptscriptstyle 1\!/\!{ }_2} \right)}^{2}}~=\text{ }[[\surd 65{{\left] /2 \right]}^{2}} \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }y\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }=\text{ }65/4 \\ 4{{x}^{2}}~+\text{ }4{{y}^{2}}~-\text{ }4y\text{ }+\text{ }1\text{ }=\text{ }65 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9c7249c8b10056630a23fd358b9d6c7a_l3.png "Rendered by QuickLaTeX.com")
∴The equation of the circle is 4x2 + 4y2 – 4y – 64 = 0 or x2 + y2 – y – 16 = 0