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)

$x^2+y^2+4x-2y+4=0$

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., $M=(\frac{k}{2}-1,1)$

Also, $MP=$ radius of circle

Thus $\sqrt{{(-2-\frac{k}{2}+1)}^2+(2-1{)}^2}=1$

$\Rightarrow \sqrt{{(-1-\frac{k}{2})}^2+1}=1$

$\Rightarrow 1+\frac{k^2}{4}+K+1=1$

$\Rightarrow k^2+4k+4=0$

$\Rightarrow (k+2{)}^2=0$

$\Rightarrow k=-2$

Therefore, mid-point $M=(-\frac{2}{2}-1,1)=(-2,1)$ which is

equal to the centre of circle.

Hence, equation of circle is

$(x+2{)}^2+(y-1{)}^2=1^2$

$\Rightarrow x^2+4x+4+y^2-2y+1=1$

$\Rightarrow x^2+y^2+4x-2y+4=0$