The points (3, 7), (5, 5)
The line x – 4y = 1…. (1)
the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2)
substituting the centre (-a, -b) in equation (1) we get,
![\[\begin{array}{*{35}{l}} \left( -\text{ }a \right)\text{ }-\text{ }4\left( -\text{ }b \right)\text{ }=\text{ }1 \\ -a\text{ }+\text{ }4b\text{ }=\text{ }1 \\ a\text{ }-\text{ }4b\text{ }+\text{ }1\text{ }=\text{ }0\ldots \ldots \text{ }\left( 3 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2aa7804c79dddb934ada8c6b1f449703_l3.png "Rendered by QuickLaTeX.com")
Substituting the points (3, 7) in equation (2), we get
![\[\begin{array}{*{35}{l}} {{3}^{2}}~+\text{ }{{7}^{2}}~+\text{ }2a\left( 3 \right)\text{ }+\text{ }2b\left( 7 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 9\text{ }+\text{ }49\text{ }+\text{ }6a\text{ }+\text{ }14b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 6a\text{ }+\text{ }14b\text{ }+\text{ }c\text{ }+\text{ }58\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c5e2faab68332979c577413ae9e8b6b_l3.png "Rendered by QuickLaTeX.com")
Substituting the points (5, 5) in equation (2), we get
![\[\begin{array}{*{35}{l}} {{5}^{2}}~+\text{ }{{5}^{2}}~+\text{ }2a\left( 5 \right)\text{ }+\text{ }2b\left( 5 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\ 25\text{ }+\text{ }25\text{ }+\text{ }10a\text{ }+\text{ }10b\text{ }+\text{ }c\text{ }=\text{ }0 \\ 10a\text{ }+\text{ }10b\text{ }+\text{ }c\text{ }+\text{ }50\text{ }=\text{ }0\ldots ..\text{ }\left( 5 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0eff08e29a22b1ac266705113d397c3d_l3.png "Rendered by QuickLaTeX.com")
By simplifying equations (3), (4) and (5) we get,
a = 3, b = 1, c = – 90
Now, by substituting the values of a, b, c in equation (2), we get
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( 3 \right)x\text{ }+\text{ }2\left( 1 \right)y\text{ }\text{ }90\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }6x\text{ }+\text{ }2y\text{ }-\text{ }90\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-671b79036b41b14774ba09df3824320f_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 + 6x + 2y – 90 = 0.