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$x^2+y^2–8x+10y–12=0$, find the centre and radius of the circles.

$\begin {array} {1 1} (A)\;(4, -5) \: and \: \sqrt{53} \\ (B)\;(-4, -5) \: and \: \sqrt{53} \\ (C)\;(-4, 5) \: and \: \sqrt{53} \\ (D)\;(4, 5) \: and \: \sqrt{53} \end {array}$

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• Equation of a circle with centre (h,k) and radius r is given as $(x-h)^2+(y-k)^2=r^2$
Equation of the given circle is
$x^2+y^2-8x+10y-12=0$
This can be rewritten as
$(x^2-8x)+(y^2+10y)=12$
(i.e) $[ (x-4)^2-16]+[(y+5)^2-25]=12$
(i.e) $(x-4)^2+(y+5)^2=53$
This is of the form
$(x-h)^2+(y-k)^2=r^2$
Comparing both the equations we get,
$h=4 , k =-5$ and $r = \sqrt{53}$
Hence the coordinates of the centre is
(4, -5) and $\sqrt{53}$
answered Jul 13, 2014