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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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$x^2+ y^2– 4x–8y–45=0$ find the centre and radius of the circles.

$\begin {array} {1 1} (A)\;(2,4) \: and \: \sqrt{65} & \quad (B)\;(-2,4) \: and \: \sqrt{65} \\ (C)\;(-2,-4) \: and \: \sqrt{65} & \quad (D)\;(2,-4) \: and \: \sqrt{65} \end {array}$

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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 circle is $x^2+y^2-4x-8y-45=0$
This can be written as
$(x^2-4x)+(y^2-8y)=45$
(i.e) $[(x-2)^2-4] +[(y-4)^2-16]=45$
(i.e) $(x-2)^2+(y-4)^2=65$
This is of the form
$(x-h)^2+(y-k)^2=r^2$
Comparing both the equations we get,
h=2 and k = 4 and r = $\sqrt{65}$ .
Thus the centre of the given circle is (2,4) and radius is $\sqrt{65}$
answered Jul 11, 2014 by thanvigandhi_1
 

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