The line 3x + 4y = 12
The value of x is 0 on meeting the y – axis. So,
![\[\begin{array}{*{35}{l}} 3\left( 0 \right)\text{ }+\text{ }4y\text{ }=\text{ }12 \\ 4y\text{ }=\text{ }12 \\ y\text{ }=\text{ }3 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d085fcd215f5f27a85019a8154c4bdde_l3.png "Rendered by QuickLaTeX.com")
The point is A(0, 3)
The value of y is 0 on meeting the x – axis. So,
![\[\begin{array}{*{35}{l}} 3x\text{ }+\text{ }4\left( 0 \right)\text{ }=\text{ }12 \\ 3x\text{ }=\text{ }12 \\ x\text{ }=\text{ }4 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a427260969256eef47519f1e97b5b689_l3.png "Rendered by QuickLaTeX.com")
The point is B(4, 0)
Since the circle passes through origin and A and B
So, AB is the diameter
the equation of the circle with AB as diameter is given by
![\[\begin{array}{*{35}{l}} (x\text{ }-\text{ }{{x}_{1}})\text{ }(x\text{ }-\text{ }{{x}_{2}})\text{ }+\text{ }(y\text{ }-\text{ }{{y}_{1}})\text{ }(y\text{ }-\text{ }{{y}_{2}})\text{ }=\text{ }0 \\ \left( x\text{ }-\text{ }0 \right)\text{ }\left( x\text{ }-\text{ }4 \right)\text{ }+\text{ }\left( y\text{ }-\text{ }3 \right)\text{ }\left( y\text{ }-\text{ }0 \right)\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }4x\text{ }-\text{ }3y\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-96729e16aa0b14facfa3c70eb6e37207_l3.png "Rendered by QuickLaTeX.com")
∴The equation of the circle is x2 + y2 – 4x – 3y = 0