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Question -

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



Answer -

Given:

x2 + y2 +6x – 14y – 1 = 0…. (1)

So the centre =[(-6/2), -(-14/2)]

= [-3, 7]

x2 + y2 –4x + 10y – 2 = 0… (2)

So the centre =[-(-4/2), (-10/2)]

= [2, -5]

We know that theequation of the circle is given by,

(x – x1) (x– x2) + (y – y1) (y – y2) = 0

(x + 3) (x – 2) + (y –7) (y + 5) = 0

Upon simplification weget

x2 +3x – 2x – 6 + y2 – 7y + 5y – 35 = 0

x2 + y2 +x – 2y – 41 = 0

The equation of thecircle is x2 + y2 + x – 2y – 41 = 0

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