Since, diameters of a circle intersect at the centre of a circle,
![\[\begin{array}{*{35}{l}} 2x\text{ }-\text{ }3y\text{ }=\text{ }5\text{ }\ldots \ldots \ldots 1 \\ 3x\text{ }-\text{ }4y\text{ }=\text{ }7\text{ }\ldots \ldots \ldots ..2 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9745eefe1c131c76382c338e3f2d520c_l3.png "Rendered by QuickLaTeX.com")
Solving the above equations,
Multiplying equation 1 by 3 we get
![\[6x\text{ }-\text{ }9y\text{ }=\text{ }15\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5d01a15ff7897acabac737fbf75e8b69_l3.png "Rendered by QuickLaTeX.com")
Multiplying equation 2 by 2 we get
![\[\begin{array}{*{35}{l}} 6x\text{ }-\text{ }8y\text{ }=\text{ }14 \\ y\text{ }=\text{ }1 \\ y\text{ }=\text{ }-1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7f77ae652a493b15cf2a2e75583ac497_l3.png "Rendered by QuickLaTeX.com")
Putting y = -1, in equation 1, we get
![\[\begin{array}{*{35}{l}} 2x\text{ }-\text{ }3\left( -1 \right)\text{ }=\text{ }5 \\ 2x\text{ }+\text{ }3\text{ }=\text{ }5 \\ 2x\text{ }=\text{ }2 \\ x\text{ }=\text{ }1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ce8b8d3eb42daa5adc6298b642ab0ce8_l3.png "Rendered by QuickLaTeX.com")
Coordinates of centre = (1,-1)
Given area = 154
Area =
![\[\begin{array}{*{35}{l}} \pi {{r}^{2}}~=\text{ }154 \\ 22/7\text{ }\times \text{ }{{r}^{2}}~=\text{ }154 \\ {{r}^{2}}~=\text{ }154\text{ }\times \text{ }7/22 \\ r\text{ }=\text{ }7\text{ }units \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e765928c3207325d9fba8d4380de5898_l3.png "Rendered by QuickLaTeX.com")
Since, the equation of a circle having centre (h, k), having radius as r units, is
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }h \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }k \right)}^{2}}~=\text{ }{{r}^{2}} \\ {{\left( x\text{ }-\text{ }1 \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{ }-\left( -1 \right) \right)}^{2}}~=\text{ }{{7}^{2}} \\ {{x}^{2}}~\text{ }-2x\text{ }+1\text{ }+\text{ }{{\left( y\text{ }+\text{ }1 \right)}^{2}}~=\text{ }49 \\ {{x}^{2}}~\text{ }-2x\text{ }+\text{ }1\text{ }+\text{ }{{y}^{2}}~+\text{ }2y\text{ }+\text{ }1\text{ }\text{ }-49\text{ }=\text{ }0 \\ {{x}^{2}}~\text{ }-2x\text{ }+\text{ }{{y}^{2}}~+\text{ }2y\text{ }\text{ }-47\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-21690a854e0b42f7e90baf06fb5ac038_l3.png "Rendered by QuickLaTeX.com")
Hence the required equation of the given circle is x2 – 2x + y2 + 2y – 47 = 0.