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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle with centre $ \bigg( \large\frac{1}{2}$$, \large\frac{1}{4} \bigg)$ and radius $ \large\frac{1}{12}$

$\begin {array} {1 1} (A)\;36x^2+36y^2-36x-18y+11=0& \quad (B)\;36x^2+36y^2+36x-18y+11=0 \\ (C)\;36x^2+36y^2-36x-18y-11=0 & \quad (D)\;36x^2+36y^2+6x-18y-11=0 \end {array}$

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  • Equation of a circle with centre (h, k) and radius r is given as : $(x-h)^2+(y-k)^2=r^2$
The coordinates of the circle are $ \bigg( \large\frac{1}{2}$$, \large\frac{1}{4} \bigg)$ and radius is $ \large\frac{1}{2}$
$ \therefore $ Equation of the circle is
$ \bigg( x - \large\frac{1}{2} \bigg)^2 $$+\bigg( y - \large\frac{1}{4} \bigg)^2$$ = \large\frac{1}{144}$
On expanding we get,
$x^2-x+ \large\frac{1}{4}$$ +y^2-\large\frac{y}{2}$$+ \large\frac{1}{16}$$= \large\frac{1}{144}$
(i.e) $x^2-x+ \large\frac{1}{4}$$+y^2-\large\frac{y}{2}$$+\large\frac{1}{16}$$-\large\frac{1}{144}$$=0$
(i.e) $144x^2+144y^2-144x-72y+44=0$
dividing throughout by 4.
$36x^2 +36y^2-36x-18y+11=0$
is the required equation of the circle.
answered Jul 11, 2014 by thanvigandhi_1
 
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