A circle passes through (0, 0) and (1, 0) and | vedclass

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(D) Let the centre of the circle be $(h, k)$. Since the circle passes through $(0, 0)$ and $(1, 0)$,the perpendicular bisector of the chord joining $(

Vedclass · Accepted Answer

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

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(D) Let the centre of the circle be $(h, k)$. Since the circle 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 $(1/2, 0)$ and the line is $x = 1/2$. Thus,$h = 1/2$.Since the circle passes through $(0, 0)$,its radius $r$ is the distance from $(1/2, k)$ to $(0, 0)$,so $r^2 = (1/2)^2 + k^2 = 1/4 + k^2$.The circle touches $x^2 + y^2 = 9$,which has centre $(0, 0)$ and radius $R = 3$. The distance between the centres is $\sqrt{(1/2)^2 + k^2} = \sqrt{1/4 + k^2} = r$.For the circles to touch,the distance between centres must be $|R \pm r|$. Here,the distance between centres is $r$,so $r = |3 \pm r|$.Case $1$: $r = 3 - r$ $\Rightarrow 2r = 3$ $\Rightarrow r = 3/2$.Then $r^2 = 9/4$,so $1/4 + k^2 = 9/4$ $\Rightarrow k^2 = 8/4 = 2$ $\Rightarrow k = \pm \sqrt{2}$.Thus,the centre is $\left( \frac{1}{2}, \pm \sqrt{2} \right)$.

$A$ circle passes through $(0, 0)$ and $(1, 0)$ and touches the circle ${x^2} + {y^2} = 9$. Find the centre of the circle.

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