x2 + y2 + 6x – 14y – 1 = 0…. (1)
So the centre
![\[\begin{array}{*{35}{l}} =\text{ }\left[ \left( -6/2 \right),\text{ }-\left( -14/2 \right) \right] \\ =\text{ }\left[ -3,\text{ }7 \right] \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }4x\text{ }+\text{ }10y\text{ }-\text{ }2\text{ }=\text{ }0\ldots \text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d494c77d3652bda39fafb758e157caae_l3.png "Rendered by QuickLaTeX.com")
So the centre = [-(-4/2), (-10/2)]
= [2, -5]
We know that the equation of the circle is given by,
![\[\begin{array}{*{35}{l}} (x\text{ }-\text{ }{{x}_{1}})\text{ }(x\text{ }-\text{ }{{x}_{2}})\text{ }+\text{ }(y\text{ }-\text{ }{{y}_{1}})\text{ }(y\text{ }-\text{ }{{y}_{2}})\text{ }=\text{ }0 \\ \left( x\text{ }+\text{ }3 \right)\text{ }\left( x\text{ }-\text{ }2 \right)\text{ }+\text{ }\left( y\text{ }-\text{ }7 \right)\text{ }\left( y\text{ }+\text{ }5 \right)\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c0a7f98ca5b9d09279c5266c251f4491_l3.png "Rendered by QuickLaTeX.com")
Upon simplification we get
![\[\begin{array}{*{35}{l}} {{x}^{2}}~+\text{ }3x\text{ }-\text{ }2x\text{ }-\text{ }6\text{ }+\text{ }{{y}^{2}}~-\text{ }7y\text{ }+\text{ }5y\text{ }-\text{ }35\text{ }=\text{ }0 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }x\text{ }-\text{ }2y\text{ }-\text{ }41\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bfb1a7d94804e9fe5b60f7dac085c8d3_l3.png "Rendered by QuickLaTeX.com")
∴The equation of the circle is x2 + y2 + x – 2y – 41 = 0
Find the equation of the circle the end points of whose diameter are the centres of the circles x^2 + y^2 + 6x – 14y – 1 = 0 and x^2 + y^2 – 4x + 10y – 2 = 0.