Question

If $(-2,2)$ and $(k,0)$ are two diametrically opposite points of a circle of radius $1$, then the equation of the circle is

• A. $x^2+y^2+2x-4y+4=0$
• B. $x^2+y^2+4x-2y-4=0$
• C. $x^2+y^2-4x+2y+4=0$
• D. $x^2+y^2+4x-2y+4=0$
• E. $x^2+y^2-4x-2y-4=0$

Answer 1

Answer: (D)

Since, $P(-2,2)$ and $Q(k,0)$ is the end points of a diameter.
$\therefore$ Mid-point $M$ of $PQ=\frac{-2+k}{2},\frac{2+0}{2}$
i.e., $M=\left(\frac{k}{2}-1,1\right)$
Also, $MP=$ radius of circle
$\therefore \sqrt{{\left(-2-\frac{k}{2}+1\right)}^2+(2-1{)}^2}=1$
$\Rightarrow \sqrt{{\left(-1-\frac{k}{2}\right)}^2+1}=1$
$\Rightarrow 1+\frac{k^2}{4}+k+1=1$
$\Rightarrow k^2+4k+4=0$
$\Rightarrow (k+2{)}^2=0$
$\Rightarrow k=-2$
$\therefore$ Mid-point $M=\left(-\frac{2}{2}-1,1\right)=(-2,1)$
which is equal to the centre of circle.
$\therefore$ Equation of circle is
$(x+2{)}^2+(y-1{)}^2=1^2$
$\Rightarrow x^2+4x+4+y^2-2y+1=1$
$\Rightarrow x^2+y^2+4x-2y+4=0$