The parabola `y^2=4x` and the circle having its center at 6, 5) intersect at right angle. Then find the possible points of intersection of these curves.

Options

A. (9,6)
B. `(2,sqrt(8))`
C. (4,4)
D. `(3,2sqrt(3))`

Explanation

To find the points of intersection of the parabola \(y^2 = 4x\) and a circle centered at \((6, 5)\) that intersect at right angles, we can follow these steps:

### Step 1: Understand the equations
The equation of the parabola is given as:
\[ y^2 = 4x \]
The general equation of a circle centered at \((h, k)\) with radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For our circle centered at \((6, 5)\), the equation becomes:
\[ (x - 6)^2 + (y - 5)^2 = r^2 \]

### Step 2: Find the tangent to the parabola
The equation of the tangent to the parabola \(y^2 = 4x\) at a point \((x_0, y_0)\) is given by:
\[ y = mx + \frac{1}{m} \]
where \(m\) is the slope of the tangent.

### Step 3: Find the point of intersection
Let the point of intersection be \((x_0, y_0)\). From the parabola, we can express \(x_0\) in terms of \(y_0\):
\[ x_0 = \frac{y_0^2}{4} \]
Substituting this into the tangent equation gives:
\[ y_0 = m\left(\frac{y_0^2}{4}\right) + \frac{1}{m} \]

### Step 4: Circle's normal condition
Since the curves intersect at right angles, the tangent to the parabola at the point of intersection must be normal to the radius of the circle at that point. The normal to the circle at \((6, 5)\) passes through the center of the circle.

### Step 5: Substitute the point into the circle's equation
Substituting \((x_0, y_0)\) into the circle's equation gives:
\[ \left(\frac{y_0^2}{4} - 6\right)^2 + (y_0 - 5)^2 = r^2 \]

### Step 6: Set up the equations
Now we have two equations:
1. The equation of the tangent line to the parabola at \((x_0, y_0)\).
2. The equation of the circle.

### Step 7: Solve for \(m\)
We can express \(y_0\) in terms of \(m\) and substitute back into the circle's equation to find \(m\). The condition for the tangents to be perpendicular gives:
\[ m_1 \cdot m_2 = -1 \]
where \(m_1\) is the slope of the tangent to the parabola and \(m_2\) is the slope of the radius from the center of the circle to the point of intersection.

### Step 8: Solve the quadratic equation
After substituting and simplifying, we will arrive at a quadratic equation in \(m\). Solving this will give us the slopes of the tangents.

### Step 9: Find points of intersection
Using the values of \(m\) found, substitute back to find the corresponding points of intersection \((x_0, y_0)\).

### Final Points of Intersection
The final points of intersection will be:
1. \((4, 2)\)
2. \((4, -2)\)