Find the equation of a circle with centre (2, | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The centre of the circle is given as $(h, k) = (2, 2)$.Since the circle passes through the point $(4, 5)$,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 the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$.$(x - 2)^{2} + (y - 2)^{2} = (\sqrt{13})^{2}$.$x^{2} - 4x + 4 + y^{2} - 4y + 4 = 13$.$x^{2} + y^{2} - 4x - 4y + 8 = 13$.$x^{2} + y^{2} - 4x - 4y - 5 = 0$.

Find the equation of a circle with centre $(2, 2)$ and passing through the point $(4, 5)$.

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