Question

The equation of the circle whose centre is on the line $2x-y-2=0$ and passes through the points $(3,-2)$ and $(-2,0)$ is

• A. $x^2+y^2-x+2y=6$
• B. $x^2+y^2-x-2y=13$
• C. $x^2+y^2-6x+2y=9$
• D. $x^2+y^2-3x+2y=0$

Answer 1

The correct option is A $x^2+y^2-x+2y=6$

Let equation of circle be
$(x-h{)}^2+(y-k{)}^2=r^2$
Putting the points $(3,-2)$ and $(-2,0)$, we get
$(3-h{)}^2+(-2-k{)}^2=r^2\dots \left(1\right)$
$(-2-h{)}^2+(0-k{)}^2=r^2\dots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$(3-h{)}^2+(-2-k{)}^2=(-2-h{)}^2+(0-k{)}^2\Rightarrow 9-6h+4+4k=4+4h\Rightarrow 10h-4k-9=0\dots \left(3\right)$

Also, the centre lies on line $2x-y-2=0$, so
$2h-k=2\dots \left(4\right)$
From equation $\left(3\right)$ and $\left(4\right)$, we get
$(h,k)=(\frac{1}{2},-1)$

From equation $\left(2\right)$,
${(-2-\frac{1}{2})}^2+(1{)}^2=r^2\Rightarrow r^2=\frac{29}{4}\Rightarrow r=\frac{\sqrt{29}}{2}$

Hence, the equation of circle is
${(x-\frac{1}{2})}^2+(y+1{)}^2=\frac{29}{4}$
$\therefore x^2+y^2-x+2y=6$