To determine the center of the circle that passes through the points \((0,0)\) and \((1,0)\) and touches the circle \(x^2 + y^2 = 9\), follow the steps below:
- Consider the general equation of a circle that passes through points \((0,0)\) and \((1,0)\). Let the center of this circle be \((h, k)\) and its radius be \(r\). Then, the equation of the circle is: \((x - h)^2 + (y - k)^2 = r^2\).
- Since the circle passes through \((0,0)\), substituting in the equation gives: \(h^2 + k^2 = r^2\).
- Similarly, since the circle also passes through \((1,0)\), substituting gives: \((1 - h)^2 + k^2 = r^2\).
- After simplifying, the equation becomes: \(1 - 2h + h^2 + k^2 = r^2\).
- From the two equations for radius we have: \(h^2 + k^2 = r^2\) and \(1 - 2h + h^2 + k^2 = r^2\). By subtracting these, we find: \(1 - 2h = 0\), which gives \(h = \frac{1}{2}\).
- The given circle \(x^2 + y^2 = 9\) has a radius of 3, centered at the origin \((0,0)\). The new circle is tangent to this, hence: \(\sqrt{h^2 + k^2} + r = 3\).
- Substituting \(h = \frac{1}{2}\) in: \(\sqrt{\left(\frac{1}{2}\right)^2 + k^2} + r = 3\).
- Using the condition \(h^2 + k^2 = r^2\): \(r = \sqrt{\left(\frac{1}{2}\right)^2 + k^2}\), thus: \(\sqrt{\frac{1}{4} + k^2} + \sqrt{\frac{1}{4} + k^2} = 3\).
- Solving for \(k\), we find \(k = \pm \sqrt{2}\).
Therefore, the center of the circle is \(\left(\frac{1}{2}, \pm \sqrt{2}\right)\).