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Equation of Circle Whose Extremities of a Diameter Given

The centre of...

Question

The centre of the circles $x^{2} + y^{2} - 6 x - 8 y - 7 = 0$ and $x^{2} + y^{2} - 4 x - 10 y - 3 = 0$ are the ends of the diameter of the circle

A

x^{2} + y^{2} - 5 x - 9 y + 26 = 0

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B

x^{2} + y^{2} + 5 x - 9 y + 14 = 0

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C

x^{2} + y^{2} + 5 x - y - 14 = 0

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D

x^{2} + y^{2} + 5 x + y + 14 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 5 x - 9 y + 26 = 0$
Centre of $(x^{2} + y^{2} - 6 x - 8 y - 7 = 0) = (3, 4)$

Centre of $(x^{2} + y^{2} - 4 x - 10 y - 3 = 0) = (2, 5)$

Equation of circle with $(3, 4)$ & $(2, 5)$ as ends of diameter

$\Rightarrow (y - 4) (x - 3) + (y - 5) (x - 2) = 0$

$\Rightarrow y^{2} - 9 y + 20 + x^{2} - 5 x + 6 = 0$

$x^{2} + y^{2} - 5 x - 9 y + 26 = 0$

$∴$ option $A$ is correct

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Equation of Circle Whose Extremities of a Diameter Given

Standard XII Mathematics

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