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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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$(x+5)^2+(y-3)^2=36$ find the centre and radius of the circles.

$\begin {array} {1 1} (A)\;(5,-3) \: and\: 6 & \quad (B)\;(-5,3) \: and\: 6 \\ (C)\;(-5,-3)\: and\: 6 & \quad (D)\;(5,3)\: and \: 6 \end {array}$

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Toolbox:
  • Equation of a circle with centre (h, k) and radius r is given as : $(x-h)^2+(y-k)^2=r^2$
The given equation of the circle is
$(x+5)^2+(y-3)^2=36$
We can rewrite the above equation as
$[x-(-5)]^2+(y-3)^2=6^2$
This is of the form
$(x-h)^2+(y-k)^2=r^2$
Here $h = -5$ and $k=3$ and $r=6$
hence the centre and radius are (-5, 3) and 6 respectively.
answered Jul 11, 2014 by thanvigandhi_1
 
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