If the equations of the two diameters of a circle are $x - y = 5$ and $2 x + y = 4$ and the radius of the circle is 5, find the equation of the circle.

x^{2} + y^{2} + 6 x - 4 y + 8 = 0

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x^{2} + y^{2} - 6 x + 4 y - 12 = 0

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x^{2} + y^{2} + 10 x + 8 = 0

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x^{2} + y^{2} - 10 y + 8 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$
Let the diameters of the circle be AB and LM whose equations are respectively
$x - y = 5$ .........(1)
$2 x + y = 4$ ..........(2)

Eq 1 + Eq 2, we get,
$x = 3$ and $y = - 2$

Since the point of intersection of any two diameters of a circle is its center. Therefore, coordinates of the center of the required circle are $(3, - 2)$ and its radius is 5.

Hence, its equation is $(x - 3)^{2} + (y + 2)^{2} = 5^{2}$
$x^{2} + y^{2} - 6 x + 4 y - 12 = 0$

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Similar questions

Find the equation of director circle of the circle whose diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area is 154 square units.

(1, - 5)

3 x + 4 y = 8,

S_{1} \equiv x^{2} + y^{2} - 4 y - 5 = 0

S_{2} \equiv x^{2} + y^{2} + 12 x + 4 y + 31 = 0

S_{2} \equiv x^{2} + y^{2} + 6 x + 12 y + 36 = 0

(1, - 2)

x^{2} + y^{2} = 1

(3, 2)

(1, 1)

x^{2} + y^{2} = 6

x^{2} + y^{2} - 6 x + 8 = 0

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