(iii) equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1)
Substituting the point (5, -8) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{5}^{2}}~+\text{ }{{\left( -\text{ }8 \right)}^{2}}~+\text{ }2a\left( 5 \right)\text{ }+\text{ }2b\left( -\text{ }8 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 25\text{ }+\text{ }64\text{ }+\text{ }10a\text{ }-\text{ }16b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 10a\text{ }-\text{ }16b\text{ }+\text{ }c\text{ }+\text{ }89\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cc84458da73b688280e71a70defb01df_l3.png "Rendered by QuickLaTeX.com")
Substituting the points (-2, 9) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{\left( -\text{ }2 \right)}^{2}}~+\text{ }{{9}^{2}}~+\text{ }2a\left( -\text{ }2 \right)\text{ }+\text{ }2b\left( 9 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 4\text{ }+\text{ }81\text{ }\text{ }4a\text{ }+\text{ }18b\text{ }+\text{ }c\text{ }=\text{ }0 \\ -4a\text{ }+\text{ }18b\text{ }+\text{ }c\text{ }+\text{ }85\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17914705b1f6380dc5e0e1dda24519b8_l3.png "Rendered by QuickLaTeX.com")
Substituting the points (2, 1) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{2}^{2}}~+\text{ }{{1}^{2}}~+\text{ }2a\left( 2 \right)\text{ }+\text{ }2b\left( 1 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 4\text{ }+\text{ }1\text{ }+\text{ }4a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 4a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }+\text{ }5\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a6f458d6fe42f44ff1c548c1a2fb9ec7_l3.png "Rendered by QuickLaTeX.com")
By simplifying equations (2), (3), (4) we get
a = 58, b = 24, c = – 285.
by substituting the values of a, b, c in equation (1), we get
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( 58 \right)x\text{ }+\text{ }2\left( 24 \right)\text{ }-\text{ }285\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }116x\text{ }+\text{ }48y\text{ }-\text{ }285\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-686f786d33b27274d47ab5c18ec37e10_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 + 116x + 48y – 285 = 0
(iv) the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1)
Substituting the points (0, 0) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{0}^{2}}~+\text{ }{{0}^{2}}~+\text{ }2a\left( 0 \right)\text{ }+\text{ }2b\left( 0 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 0\text{ }+\text{ }0\text{ }+\text{ }0a\text{ }+\text{ }0b\text{ }+\text{ }c\text{ }=\text{ }0 \\ c\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fdfb4b7bbb1cd3b1059c04dda26e0cc1_l3.png "Rendered by QuickLaTeX.com")
Substituting the points (-2, 1) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{\left( -\text{ }2 \right)}^{2}}~+\text{ }{{1}^{2}}~+\text{ }2a\left( -\text{ }2 \right)\text{ }+\text{ }2b\left( 1 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 4\text{ }+\text{ }1\text{ }\text{ }4a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\ -4a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }+\text{ }5\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3f2c1798c5ab2f83ecbe2fe53c94feea_l3.png "Rendered by QuickLaTeX.com")
Substitute the points (-3, 2) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{\left( -\text{ }3 \right)}^{2}}~+\text{ }{{2}^{2}}~+\text{ }2a\left( -\text{ }3 \right)\text{ }+\text{ }2b\left( 2 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 9\text{ }+\text{ }4\text{ }-\text{ }6a\text{ }+\text{ }4b\text{ }+\text{ }c\text{ }=\text{ }0 \\ -6a\text{ }+\text{ }4b\text{ }+\text{ }c\text{ }+\text{ }13\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-85ebe19474314f39823579cb84fccc4f_l3.png "Rendered by QuickLaTeX.com")
By simplifying the equations (2), (3), (4) we get
a = -3/2, b = -11/2, c = 0
by substituting the values of a, b, c in equation (1), we get
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( -3/2 \right)x\text{ }+\text{ }2\left( -11/2 \right)y\text{ }+\text{ }0\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }3x\text{ }-\text{ }11y\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c964f6a412def92b089c577cfdf407d_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 – 3x – 11y = 0