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(B) Let the required circle be $S_2$ with centre $(h, k)$ and radius $r$.Since it passes through $(0, 0)$ and $(1, 0)$,the perpendicular bisector of t

Vedclass · Accepted Answer

$\left( \frac{1}{2}, -\sqrt{2} \right)$

Vedclass · Answer

(B) Let the required circle be $S_2$ with centre $(h, k)$ and radius $r$.Since it passes through $(0, 0)$ and $(1, 0)$,the perpendicular bisector of the chord joining $(0, 0)$ and $(1, 0)$ must contain the centre. The midpoint is $(\frac{1}{2}, 0)$,so $h = \frac{1}{2}$.The radius $r$ is the distance from $(\frac{1}{2}, k)$ to $(0, 0)$,so $r^2 = (\frac{1}{2})^2 + k^2 = \frac{1}{4} + k^2$.The circle $S_2$ touches $x^2 + y^2 = 9$ (centre $(0, 0)$,radius $R = 3$).For internal contact,the distance between centres $d = R - r$.$d^2 = (\frac{1}{2})^2 + k^2 = r^2$.So $r^2 = (3 - r)^2 = 9 - 6r + r^2$,which gives $6r = 9$,so $r = \frac{3}{2}$.Substituting $r^2 = \frac{9}{4}$ into $r^2 = \frac{1}{4} + k^2$ gives $\frac{9}{4} = \frac{1}{4} + k^2$,so $k^2 = 2$,$k = \pm \sqrt{2}$.Thus,the centre is $\left( \frac{1}{2}, \pm \sqrt{2} \right)$. Given the options,the correct choice is $\left( \frac{1}{2}, -\sqrt{2} \right)$.

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

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