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Find the equation of the circle with centre $(1,1)$ and radius $\sqrt 2$

$\begin{array}{1 1} x^2 +y^2 - 2x-2y = 0\\ x^2 +y^2 - 2x+2y = 0 \\ x^2 +y^2 + 2x-2y = 0 \\ x^2 +y^2 + 2x+2y = 0 \end{array} $

1 Answer

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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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Given: circle with centre $(1,1)$ and radius $\sqrt 2$, $\rightarrow r = \sqrt 2, h = 1, k = 1$
Therefore equation of circle using the distance formula is: $ (x-1)^2 + (y-1)^2 = (\sqrt 2)^2$
$\quad \quad x^2-2x+1+y^2-2y+1 = 2$
$\quad \quad x^2 +y^2 - 2x-2y = 0$
answered Apr 1, 2014 by balaji.thirumalai
 

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