Question

The centre of circle passing through (0, 0) and (1, 0) and touching the circle $x^2+y^2=9$ is
(2002)

• A. $\left(\begin{array}{c}\frac{1}{2},\frac{1}{2}\end{array}\right)$
• B. $\left(\begin{array}{c}\frac{1}{2},\sqrt{2}\end{array}\right)$
• C. $\left(\begin{array}{c}\frac{3}{1},\frac{1}{2}\end{array}\right)$
• D. $\left(\begin{array}{c}\frac{1}{2},\frac{3}{2}\end{array}\right)$

Answer 1

The correct option is B

$\left(\begin{array}{c}\frac{1}{2},\sqrt{2}\end{array}\right)$

Let the required circle be
$x^2+y^2+2gx+2fy+c=0$
Since it passes through (0,0) and (1, 0)
$\Rightarrow c=0andg=\frac{1}{2}$

Points (0, 0) and (1, 0) lie inside the circle $x^2+y^2=9,$ so two circles touch internally
$\Rightarrow c_1{c}_2=r_1-r_2$
$\therefore \sqrt{g^2+f^2}=3-\sqrt{g^2+f^2}\Rightarrow \sqrt{g^2+f^2}=\frac{3}{2}$
$\Rightarrow f^2=\frac{9}{4}-\frac{1}{4}=2\therefore f=\pm \sqrt{2}.$
Hence, the centres of required circle are
$\left(\begin{array}{c}\frac{1}{2},\sqrt{2}\end{array}\right)$ or
$\left(\begin{array}{c}\frac{1}{2},-\sqrt{2}\end{array}\right)$