(i) (5, 7), (8, 1) and (1, 3)
By using the standard form of the equation of the circle:
x2 + y2 + 2ax + 2by + c = 0 ….. (1)
Substitute the given point (5, 7) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{5}^{2}}~+\text{ }{{7}^{2}}~+\text{ }2a\text{ }\left( 5 \right)\text{ }+\text{ }2b\text{ }\left( 7 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 25\text{ }+\text{ }49\text{ }+\text{ }10a\text{ }+\text{ }14b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 10a\text{ }+\text{ }14b\text{ }+\text{ }c\text{ }+\text{ }74\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-159800ca37ac740cda0b97748d8a463a_l3.png "Rendered by QuickLaTeX.com")
substituting the given point (8, 1) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{8}^{2}}~+\text{ }{{1}^{2}}~+\text{ }2a\text{ }\left( 8 \right)\text{ }+\text{ }2b\text{ }\left( 1 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 64\text{ }+\text{ }1\text{ }+\text{ }16a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 16a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }+\text{ }65\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11c2226b73bad74f4c402921accd323c_l3.png "Rendered by QuickLaTeX.com")
Substituting the point (1, 3) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{1}^{2}}~+\text{ }{{3}^{2}}~+\text{ }2a\text{ }\left( 1 \right)\text{ }+\text{ }2b\text{ }\left( 3 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 1\text{ }+\text{ }9\text{ }+\text{ }2a\text{ }+\text{ }6b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 2a\text{ }+\text{ }6b\text{ }+\text{ }c\text{ }+\text{ }10\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d23e44d45747e5660804f1642c80e8f_l3.png "Rendered by QuickLaTeX.com")
Now by simplifying the equations (2), (3), (4) we get the values
a = -29/6, b = -19/6, c = 56/3
Substituting the values of a, b, c in equation (1), we get
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\text{ }\left( -29/6 \right)x\text{ }+\text{ }2\text{ }\left( -19/6 \right)\text{ }+\text{ }56/3\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }29x/3\text{ }-\text{ }19y/3\text{ }+\text{ }56/3\text{ }=\text{ }0 \\ 3{{x}^{2}}~+\text{ }3{{y}^{2}}~-\text{ }29x\text{ }-\text{ }19y\text{ }+\text{ }56\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c91c5c60b1bfb07a85a870188d6384ff_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is 3x2 + 3y2 – 29x – 19y + 56 = 0
(ii) (1, 2), (3, – 4) and (5, – 6)
using the standard form of the equation of the circle:
x2 + y2 + 2ax + 2by + c = 0 ….. (1)
Substituting the points (1, 2) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{1}^{2}}~+\text{ }{{2}^{2}}~+\text{ }2a\text{ }\left( 1 \right)\text{ }+\text{ }2b\text{ }\left( 2 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 1\text{ }+\text{ }4\text{ }+\text{ }2a\text{ }+\text{ }4b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 2a\text{ }+\text{ }4b\text{ }+\text{ }c\text{ }+\text{ }5\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d5a2aee21ff189038630facbc3124292_l3.png "Rendered by QuickLaTeX.com")
Substitute the points (3, -4) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{3}^{2}}~+\text{ }{{\left( -\text{ }4 \right)}^{2}}~+\text{ }2a\text{ }\left( 3 \right)\text{ }+\text{ }2b\text{ }\left( -\text{ }4 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 9\text{ }+\text{ }16\text{ }+\text{ }6a\text{ }\text{ }8b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 6a\text{ }-\text{ }8b\text{ }+\text{ }c\text{ }+\text{ }25\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-14b068479d37abd10b91c4bb8db8bb39_l3.png "Rendered by QuickLaTeX.com")
Substituting the points (5, -6) in equation (1), we get
![\[\begin{array}{*{35}{l}} {{5}^{2}}~+\text{ }{{\left( -\text{ }6 \right)}^{2}}~+\text{ }2a\text{ }\left( 5 \right)\text{ }+\text{ }2b\text{ }\left( -\text{ }6 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 25\text{ }+\text{ }36\text{ }+\text{ }10a\text{ }-\text{ }12b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 10a\text{ }-\text{ }12b\text{ }+\text{ }c\text{ }+\text{ }61\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-760075e50981a42fbbcff9c634d94253_l3.png "Rendered by QuickLaTeX.com")
Now by simplifying the equations (2), (3), (4) we get
a = – 11, b = – 2, c = 25
Substitute the values of a, b and c in equation (1), we get
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( -\text{ }11 \right)x\text{ }+\text{ }2\left( -\text{ }2 \right)\text{ }+\text{ }25\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }22x\text{ }-\text{ }4y\text{ }+\text{ }25\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c3d90d1fb9da20e1386bf76fc1f0acf1_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 – 22x – 4y + 25 = 0