Question

The center of a circle which passes through the points $(0,0),(1,0)$ and touches the circles $x^2+y^2=9$ is:

• A. $(\frac{3}{2},\frac{1}{2})$
• B. $(\frac{1}{2},\frac{3}{2})$
• C. $(\frac{1}{2},\frac{1}{2})$
• D. $(\frac{1}{2},\sqrt{2})$

Answer 1

The correct option is A $(\frac{3}{2},\frac{1}{2})$

Let the points $P(0,0)$ and $Q(1,0)$ touches the circle $x^2+y^2=r^2$

Equation of circle is given by ${(x-h)}^2+(y-k{)}^2=r^2----------(1)$

consider point $P(0,0)$, we get

${(0-h)}^2+{(0-k)}^2=r^2$

$⟹h^2+k^2=r^2--------------------------\left(2\right)$

consider point Q(1,0), we get

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

$⟹1+h^2-2h+k^2=r^2---------------------\left(3\right)$

from equations (2) and (3), we get

$h^2+k^2=1+h^2-2h+k^2$

$⟹1-2h=0$

$⟹h=\frac{1}{2}$

Since, circle (1) touches circle (2), then the center is given by center of

circle (1)$=(h,k)$ and radius $r$

center of circle (2) $=(0,0)$ and radius 3

distance between their centers =$\sqrt{{(h-0)}^2+(k-0{)}^2}=\sqrt{{\left(h\right)}^2+(k{)}^2}⟹\pm r$

So, $-r=r-3$

$⟹2r=3$

$⟹r=\frac{3}{2}$

from equation (2),

$h^2+k^2=r^2$

$⟹{\left(\frac{1}{2}\right)}^2+k^2={\left(\frac{3}{2}\right)}^2$

$\Rightarrow k^2=\frac{9}{4}-\frac{1}{4}$ $=\frac{8}{4}=2$

$\Rightarrow k=\pm \sqrt{2}$

So, the center of circle (1) $=(h,k)=(\frac{1}{2},\sqrt{2})or(\frac{1}{2},-\sqrt{2})$

Hence, option (d) is correct.