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Find the Equation of the Circle the End Points of Whose Diameter Are the Centres of the Circles X2 + Y2 + 6x − 14y − 1 = 0 and X2 + Y2 − 4x + 10y − 2 = 0. - Mathematics

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Question

Find the equation of the circle the end points of whose diameter are the centres of the circles x2 + y2 + 6x − 14y − 1 = 0 and x2 + y2 − 4x + 10y − 2 = 0.

Solution

Given:

\[x^2 + y^2 + 6x - 14y - 1 = 0\]  ...(1)
And,
\[x^2 + y^2 - 4x + 10y - 2 = 0\]...(2)
Equations (1) and (2) can be rewritten as follows:
\[\left( x + 3 \right)^2 + \left( y - 7 \right)^2 = 59\]
And,
\[\left( x - 2 \right)^2 + \left( y + 5 \right)^2 = 31\]
Thus, the centres of the circles are (−3, 7) and (2, −5).
Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is
\[\left( x + 3 \right)\left( x - 2 \right) + \left( y - 7 \right)\left( y + 5 \right) = 0\]
\[x^2 + y^2 + x - 2y - 41 = 0\]
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Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.3 [Page 37]

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RD Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.3 | Q 2 | Page 37

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