1. Circle through $(0,0)$ and $(1,0)$

Let the required circle have centre $(h,k)$ and radius $r$.

Since it passes through $(0,0)$ and $(1,0)$, $r^2=h^2+k^2$ and also $r^2=(h-1)^2+k^2.$

Equating, $h^2+k^2=(h-1)^2+k^2$ $h^2=h^2-2h+1$ $2h=1$ $h=\frac12.$

So the centre is of the form $(h,k)=\left(\frac12,k\right).$

Also, $r^2=h^2+k^2=\frac14+k^2.$ Hence $r=\sqrt{\frac14+k^2}.$

2. Touching the circle $x^2+y^2=9$

The given circle has centre $(0,0)$ and radius $3$.

Distance between the centres is $d=\sqrt{h^2+k^2}=r,$ because $(0,0)$ lies on the required circle.

If two circles touch, then either:

• externally: $d=r+3$, or
• internally: $d=|3-r|$.

But here $d=r$, so external touching would give $r=r+3,$ which is impossible.

Therefore touching must be internal: $d=3-r.$ Since $d=r$, $r=3-r$ $2r=3$ $r=\frac32.$

3. Find $h^2+k^2$

Because $r^2=h^2+k^2$, $h^2+k^2=\left(\frac32\right)^2=\frac94.$

Therefore, $4(h^2+k^2)=4\cdot \frac94=9.$

4. Final answer

$\boxed{9}$

This value is the same for all possible centres $(h,k)$.

5. Comparison with stored answer

Stored correct answer: $9$

Our derived answer: $9$

So the derived answer agrees with the stored answer.