Question

Find the radius and centre of the circle represented by the equation:
$2x^2+2y^2-x=0$

Answer 1

Given $2x^2+2y^2-x=0$ ______(1)

we need to make this form

$(x-h{)}^2+(y-k{)}^2=r^2$

From (i)

$2x^2+2y^2-x=0$

$2(x^2+y^2-\frac{x}{2})=0$

$\Rightarrow x^2+y^2-\frac{x}{2}=\frac{0}{2}$

$\Rightarrow x^2-\frac{x}{2}+y^2=0$

$\Rightarrow (x{)}^2-2.\times .(\frac{1}{4})x+y^2=0$

$\Rightarrow .\left(x^2\right)-2.\left(\frac{1}{4}\right)x+\left(\frac{1}{4}\right)x+y^2=0$

$\Rightarrow [x^2-2.(\frac{1}{4})x+(\frac{1}{4}{)}^2]-(\frac{1}{4}{)}^2+y^2=0$

$\Rightarrow [x^2-2.(\frac{1}{4}{)}^2]-(\frac{1}{4}{)}^2+y^2=0$

Using $(a-b{)}^2=a^2-2ab+b^2$

$(x-\frac{1}{4}{)}^2-(\frac{1}{4}{)}^2=0$

$(x-\frac{1}{4}{)}^2+(y-0{)}^2=(\frac{1}{4}{)}^2$ ________ (2)

comparing (2) with $(x-b{)}^2+(y-k{)}^2=r^2$

$h=\frac{1}{4},k=0;r=\frac{1}{4}$

Hence center of circle $\left(\frac{1}{4}\&0\right)$

Radius of circle = $r=\frac{1}{4}$