Find the equation of the circle two of whose diameters are x + y = 3 a

1.	Find the equation of the circle two of whose diameters are x + y = 3 and 2x + y = 2 and passing through the centre of the circle x2 + y2 – 4x + 2y – 1 = 0.
Answer» Given x²+ y² – 4x + 2y – 1 = 0 P = (2,-1) & centre of the circle is point of intersection of diameters x + y = 3 – (1) & 2x + y = 2 – (2) eqn. (2) – eqn. (1) gives x = -1 & y = 3 – x = (3 – C – 1) = 4 ∴ centre = c(-1, 4) r = CP = \(\sqrt{(-1 - 2)^2 - (4 - (-1)^2} - \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25}\) = √34 ∴ the eqn. of the circle with c = (-1, 4) & r = √34 is (x + 1)²+ (y – 4)²= (√34)² x² + 1 + 2x + y² + 16 – 8y = 34 x² + y² + 2x – 8y – 17 = 0.

Find the equation of the circle two of whose diameters are x + y = 3 and 2x + y = 2 and passing through the centre of the circle x2 + y2 – 4x + 2y – 1 = 0.

Answer»

Given x²+ y² – 4x + 2y – 1 = 0

P = (2,-1) & centre of the circle is point of intersection of diameters

x + y = 3 – (1) & 2x + y = 2 – (2)

eqn. (2) – eqn. (1) gives x = -1 & y = 3 – x = (3 – C – 1) = 4

∴ centre = c(-1, 4)

r = CP

= \(\sqrt{(-1 - 2)^2 - (4 - (-1)^2} - \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25}\)

= √34

∴ the eqn. of the circle with c = (-1, 4) & r = √34 is

(x + 1)²+ (y – 4)²= (√34)²

x² + 1 + 2x + y² + 16 – 8y = 34

x² + y² + 2x – 8y – 17 = 0.

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Find the equation of the circle two of whose diameters are x + y = 3 and 2x + y = 2 and passing through the centre of the circle x2 + y2 – 4x + 2y – 1 = 0.

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