Question

The centre of the circle $\left(s\right)$ passing through the points $(0,0),(1,0)$ and touching the circle $x^2+y^2=9,$ is (are)

• A. $(3/2,1/2)$
• B. $(1/2,3/2)$
• C. $(1/2,2^{1/2})$
• D. $(1/2,-2^{1/2})$

Answer 1

Answer: (C), (D)

The correct options are
C $(1/2,-2^{1/2})$
D $(1/2,2^{1/2})$

Since the required circle touches the given circle and passes through
its centre $(0,0),$ its radius is half that of the given circle.

Let the equation of the required circle be

$(x-h{)}^2+(y-k{)}^2=(3/2{)}^2$

Since it passes through $(0,0)$ and $(1,0),$ we have
$h^2+k^2=9/4$ and $(h-1{)}^2+k^2=9/4$

$\Rightarrow h^2-(h-1{)}^2=0\Rightarrow 2h-1=0\Rightarrow h=1/2$
so that $(1/2{)}^2+k^2=9/4\Rightarrow =k^2=2\Rightarrow k=\pm 2^{1/2}$

Therefore, the, required centres are $(1/2,2^{1/2})$ and $(1/2,-2^{1/2})$.