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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.

x2 + y2 + 6x – 14y – 1 = 0…. (1)

So the centre

   

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

= [2, -5]

We know that the equation of the circle is given by,

   

Upon simplification we get

   

∴The equation of the circle is x2 + y2 + x – 2y – 41 = 0