(iii) By using the standard equation formula,
![\[{{\left( x\text{ }-\text{ }a \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }b \right)}^{2}}~=\text{ }{{r}^{2}}~\ldots .\text{ }\left( 1 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ac20885842b2118f9114910cb9e5db75_l3.png "Rendered by QuickLaTeX.com")
converting given circle’s equation into the standard form.
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~\text{ }4x\text{ }+\text{ }6y\text{ }=\text{ }5 \\ ({{x}^{2}}~-\text{ }4x\text{ }+\text{ }4)\text{ }+\text{ }({{y}^{2}}~+\text{ }6y\text{ }+\text{ }9)\text{ }=\text{ }5\text{ }+\text{ }4\text{ }+\text{ }9 \\ {{\left( x\text{ }-\text{ }2 \right)}^{2}}~+\text{ }{{\left( y\text{ }+\text{ }3 \right)}^{2}}~=\text{ }18 \\ {{\left( x\text{ }-\text{ }2 \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{ }-\left( -3 \right) \right)}^{2}}~=\text{ }{{\left( 3\surd 2 \right)}^{2}}~\ldots \text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8168e03e618470ade4eb3d7722984d71_l3.png "Rendered by QuickLaTeX.com")
By comparing equation (2) with (1), we get
Centre = (2, -3) and radius = 3√2
∴ The centre of the circle is (2, -3) and the radius is 3√2.
(iv)
The equation x2 + y2 – x + 2y – 3 = 0
We need to find the centre and the radius.
By using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Now let us convert given circle’s equation into the standard form.
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }x\text{ }+\text{ }2y\text{ }-\text{ }3\text{ }=\text{ }0 \\ ({{x}^{2}}~-\text{ }x\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_4})\text{ }+\text{ }({{y}^{2}}~+\text{ }2y\text{ }+\text{ }1)\text{ }-\text{ }3\text{ }-\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }-\text{ }1\text{ }=\text{ }0 \\ {{\left( x\text{ }-\text{ }{\scriptscriptstyle 1\!/\!{ }_2} \right)}^{2}}~+\text{ }{{\left( y\text{ }+\text{ }1 \right)}^{2}}~=\text{ }17/4\text{ }\ldots .\text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f22bc76ba45bd31409680d2e3aaafb62_l3.png "Rendered by QuickLaTeX.com")
By comparing equation (2) with (1), we get
Centre = (½, – 1) and radius = √17/2
∴ The centre of the circle is (½, -1) and the radius is √17/2.