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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the centre and radius of the circle $x^2+y^2-4x-9y-45=0$

$\begin{array}{1 1}centre = (2,4)\; and radius = \sqrt {65} \\ centre = (2,-4)\;and \;radius = \sqrt {65} \\ centre = (-2,4) \;and \;radius = \sqrt {65} \\ centre = (-2,-4)\; and radius = \sqrt {65} \end{array} $

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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: the equation of the circle, we need to rewrite it in the form $(x-h)^2+(y-k)^2=r^2$, so that we can determine the centre as (h,k) and radius as $r$
Given $x^2+y^2-4x-9y-45=0$
$\quad \quad x^2-2 \times 2x + 4 + y^2 - 2\times 4y + 16 - 4 - 16 = 45$
$\quad \quad (x-2)^2 + (y-4)^2 = 65$
$\Rightarrow$ centre = $(2,4)$ and radius = $\sqrt {65}$
answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai
 

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