Question

Consider a family of circle which are passing through the point $(-1,1)$ and are tangent to x-axis. If $(h,k)$ are the co-ordinates of the centre of the circles, then the set of values of $k$ is given by the interval

• A. $k\ge \frac{1}{2}$
• B. $-1\frac{1}{2}\le k\le \frac{1}{2}$
• C. $k\le \frac{1}{2}$
• D. $0<k<\frac{1}{2}$

Answer 1

The correct option is A $k\ge \frac{1}{2}$

Consider

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

This circle passes through $(-1,1)$

$(-1-h{)}^2+(1-k{)}^2=r^2{h}^2+k^2+2h+1+1-2k=r^2{h}^2+k^2+2h-2k+2-r^2=0\left(1\right)$

Point of coincidence $(x-h{)}^2+(0-k{)}^2=r^2(x-h)=\pm \sqrt{r^2-k^2}x=(h\pm \sqrt{r^2-k^2})$

Point of coincidence should be $(h,0)$ and radius of circle should be $k$

From equation $\left(1\right)$

$h^2+k^2+2h-2k+2-k^2=0h^2+2h+2-2k=0D=4-4(2-2k)=-4+8k$

For $h$ to be real $D\ge 0$ $-4+8k\ge 0k\ge \frac{1}{2}$