Browse Questions

# Find the equation of the circle with centre $\bigg( \large\frac{1}{2}$$, \large\frac{1}{4} \bigg) and radius \large\frac{1}{12} \begin {array} {1 1} (A)\;36x^2+36y^2-36x-18y+11=0& \quad (B)\;36x^2+36y^2+36x-18y+11=0 \\ (C)\;36x^2+36y^2-36x-18y-11=0 & \quad (D)\;36x^2+36y^2+6x-18y-11=0 \end {array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • Equation of a circle with centre (h, k) and radius r is given as : (x-h)^2+(y-k)^2=r^2 The coordinates of the circle are \bigg( \large\frac{1}{2}$$, \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.