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

Question

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

Accepted Answer

Let the circle be $x^2+y^2+2gx+2fy+c=0$. Through $(0,0)$: $c=0$. Through $(1,0)$: $1+2g=0\Rightarrow g=-\tfrac12$.

So the circle is $x^2+y^2-x+2fy=0$, with centre $\left(\tfrac12,-f\right)$ and radius $\sqrt{\tfrac14+f^2}$. Since the circle passes through the origin, this radius equals the distance from the centre to the origin --- automatically consistent.

Internal tangency with $x^2+y^2=9$ (radius $3$, centre at the origin): distance between centres $=3-r$, but the distance between centres is itself $r$ (as just noted), so
$$r=3-r\Rightarrow r=\frac32\ \Rightarrow\ \frac14+f^2=\frac94\Rightarrow f^2=2\Rightarrow f=\pm\sqrt2.$$

Centre: $\left(\tfrac12,-f\right)=\left(\tfrac12,\mp\sqrt2\right)$, matching option D, $\left(\tfrac12,-\sqrt2\right)$.

*(The source's recorded answer letter did not match this derivation; option D is the value consistent with the computation above.)*

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