The equation of a circle whose centre lies at | vedclass

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Question

(A) The centre of the circle is $(h, k) = (-2, 1)$.Since the circle passes through the point $(4, 3)$,the radius $r$ is the distance between the centr

Vedclass · Accepted Answer

${x^2} + {y^2} + 4x - 2y - 35 = 0$

Vedclass · Answer

(A) The centre of the circle is $(h, k) = (-2, 1)$.Since the circle passes through the point $(4, 3)$,the radius $r$ is the distance between the centre $(-2, 1)$ and the point $(4, 3)$.$r^2 = (4 - (-2))^2 + (3 - 1)^2 = (6)^2 + (2)^2 = 36 + 4 = 40$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - (-2))^2 + (y - 1)^2 = 40$.$(x + 2)^2 + (y - 1)^2 = 40$.$x^2 + 4x + 4 + y^2 - 2y + 1 = 40$.$x^2 + y^2 + 4x - 2y + 5 = 40$.$x^2 + y^2 + 4x - 2y - 35 = 0$.

The equation of a circle whose centre lies at the point $(-2, 1)$ and which passes through the point $(4, 3)$ is:

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