Find the equation of the circle the end points of whose diameter are the centre of the circles
$x^{2} + y^{2} - 6 x - 14 y - 1 = 0$
and $x^{2} + y^{2} - 4 x + l O y - 2 = 0$

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Solution

Centre of the circles
$x^{2} + y^{2} + 6 x - 14 y - 1 = 0 i s (- 3, 7)$
and $x^{2} + 9 - 4 x + l 0 y - 2 = 0$ is (2, -5)
Equation of the circle is
$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0 \Rightarrow (x + 3) (x - 2) + (y - 7) (y + 5) = 0 \Rightarrow= x^{2} + 3 x - 2 x - 6 + y^{2} - 7 y + 5 y - 35 = 0 \Rightarrow x^{2} + y^{2} + x - 2 y - 41 = 0$

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x^{2} + y^{2} - 6 x - 8 y - 7 = 0

x^{2} + y^{2} - 4 x - 10 y - 3 = 0

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The equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0, is

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

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7

x -

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x -

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