(x-2)^2+y^2=9 y=x^2-4x+4

To solve the system of equations given by:

1. \((x-2)^2 + y^2 = 9\) (this is the equation of a circle with center at (2, 0) and radius 3)
2. \(y = x^2 - 4x + 4\) (this is a quadratic equation, which can also be expressed as \(y = (x-2)^2\), a parabola with vertex at (2,0))

we'll substitute the expression for \(y\) from the second equation into the first equation.

Step 1: Substitute for \(y\)

Substituting \(y = (x-2)^2\) into the first equation:

\[ (x-2)^2 + ((x-2)^2)^2 = 9 \]

Step 2: Simplify

Let \(u = (x-2)^2\). The equation now becomes:

\[ u + u^2 = 9 \]

This can be rearranged to form a standard polynomial equation:

\[ u^2 + u - 9 = 0 \]

Step 3: Solve the quadratic

We will use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 1\), and \(c = -9\):

\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} \] \[ u = \frac{-1 \pm \sqrt{1 + 36}}{2} \] \[ u = \frac{-1 \pm \sqrt{37}}{2} \]

This gives us two possible values for \(u\):

\[ u_1 = \frac{-1 + \sqrt{37}}{2} \] \[ u_2 = \frac{-1 - \sqrt{37}}{2} \quad (\text{this value is negative and can be discarded since } u = (x-2)^2 \geq 0) \]

So we only consider:

\[ u = \frac{-1 + \sqrt{37}}{2} \]

Step 4: Find \(x\)

Recall that \(u = (x-2)^2\):

\[ (x-2)^2 = \frac{-1 + \sqrt{37}}{2} \]

Taking the square root of both sides gives:

\[ x - 2 = \pm \sqrt{\frac{-1 + \sqrt{37}}{2}} \]

So we have:

\[ x = 2 + \sqrt{\frac{-1 + \sqrt{37}}{2}} \quad \text{and} \quad x = 2 - \sqrt{\frac{-1 + \sqrt{37}}{2}} \]

Step 5: Find \(y\)

Now, substitute \(x\) back into \(y = (x-2)^2\):

For each value of \(x\):

1. \(y_1 = (2 + \sqrt{\frac{-1 + \sqrt{37}}{2}} - 2)^2 = \frac{-1 + \sqrt{37}}{2}\)
2. \(y_2 = (2 - \sqrt{\frac{-1 + \sqrt{37}}{2}} - 2)^2 = \frac{-1 + \sqrt{37}}{2}\)

Thus, the solutions to the system of equations are:

\[ \left(2 + \sqrt{\frac{-1 + \sqrt{37}}{2}}, \frac{-1 + \sqrt{37}}{2}\right) \]

and

\[ \left(2 - \sqrt{\frac{-1 + \sqrt{37}}{2}}, \frac{-1 + \sqrt{37}}{2}\right) \]

These are the intersection points of the circle and the parabola.