Find the equation of the circle the end points of whose diameter are t

1.	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.
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 the equation 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 we get x² + 3x – 2x – 6 + y² – 7y + 5y – 35 = 0 x² + y² + x – 2y – 41 = 0 ∴The equation of the circle is x² + y² + x – 2y – 41 = 0

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.

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 the equation 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 we get

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

x² + y² + x – 2y – 41 = 0

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

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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.

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