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

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  • Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
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Here we are given the centre of the circle $(h,k)$ = $(2,2)$.
Since the circle passes through the point $(4,5)$, we can calculate the radius as $r^2 = (2-4)^2 + (2-5)^2 = 2^2+(-3)^2 = 13 \rightarrow r = \sqrt {13}$
Given $h,k$ and $r$, we can write the equation of the circle:$(x-h)^2+(y-k)^2=r^2$ as
$\Rightarrow (x-2)^2+(y-2)^2 = \sqrt{13}^2 \rightarrow x^2+y^2+16-4x-4y = 13 $
$\Rightarrow x^2+y^2-4x-4y-5 = 0 $
answered Apr 1, 2014 by balaji.thirumalai

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