The equation of the circle passing through (4 | vedclass

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Question

(B) The centre of the circle is $(h, k) = (2, 2)$.Since the circle passes through $(4, 5)$,the radius $r$ is the distance between $(2, 2)$ and $(4, 5)

Vedclass · Accepted Answer

$x^2 + y^2 - 4x - 4y - 5 = 0$

Vedclass · Answer

(B) The centre of the circle is $(h, k) = (2, 2)$.Since the circle passes through $(4, 5)$,the radius $r$ is the distance between $(2, 2)$ and $(4, 5)$:$r = \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values: $(x - 2)^2 + (y - 2)^2 = (\sqrt{13})^2$.$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 13$.$x^2 + y^2 - 4x - 4y + 8 = 13$.$x^2 + y^2 - 4x - 4y - 5 = 0$.

The equation of the circle passing through $(4, 5)$ and having the centre at $(2, 2)$ is:

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