The circle `x^2 + y^2 - 2x - 6y+2=0` intersects the parabola `y^2 = 8x` orthogonally at the point `P`. The equation of the tangent to the parabola at `P` can be

Options

A. x-y-4=0
B. 2x+y-2=0
C. x+y-4=0
D. 2x-y+1=0

Explanation

To solve the problem step by step, we will first analyze the given equations of the circle and the parabola, find their points of intersection, and then determine the equation of the tangent to the parabola at the point of intersection.

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 2x - 6y + 2 = 0 \]
We can rewrite this in standard form by completing the square for both \(x\) and \(y\).

### Step 2: Complete the square for \(x\) and \(y\)
For \(x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]
Substituting back into the equation gives:
\[ (x - 1)^2 - 1 + (y - 3)^2 - 9 + 2 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y - 3)^2 - 8 = 0 \]
Thus, the circle can be expressed as:
\[ (x - 1)^2 + (y - 3)^2 = 8 \]
This indicates a circle centered at \((1, 3)\) with a radius of \(2\sqrt{2}\).

### Step 3: Write the equation of the parabola
The equation of the parabola is given as:
\[ y^2 = 8x \]

### Step 4: Find the points of intersection
To find the intersection points, we can substitute \(y^2 = 8x\) into the circle's equation. From the parabola, we can express \(x\) in terms of \(y\):
\[ x = \frac{y^2}{8} \]
Substituting this into the circle's equation:
\[ \left(\frac{y^2}{8} - 1\right)^2 + (y - 3)^2 = 8 \]
Expanding and simplifying this equation will yield a quadratic in \(y\).

### Step 5: Solve the quadratic equation
After simplifying, we will get a quadratic equation in \(y\). Solving this will give us the \(y\)-coordinates of the intersection points.

### Step 6: Find the corresponding \(x\) values
Once we have the \(y\) values, we can substitute them back into \(x = \frac{y^2}{8}\) to find the corresponding \(x\) values.

### Step 7: Determine the slope of the tangent line
The slope \(m\) of the tangent line to the parabola \(y^2 = 8x\) at any point \((x_0, y_0)\) can be found using the derivative:
\[ \frac{dy}{dx} = \frac{4}{y} \]
At the intersection point \(P\), we will substitute the \(y\) value to find the slope \(m\).

### Step 8: Write the equation of the tangent line
Using the point-slope form of the equation of a line, the equation of the tangent line at point \(P\) can be expressed as:
\[ y - y_0 = m(x - x_0) \]

### Step 9: Check for orthogonality
To check if the circle and the parabola intersect orthogonally at point \(P\), we need to verify that the product of the slopes of the tangent lines at point \(P\) is \(-1\).

### Step 10: Finalize the tangent equations
From the calculations, we will find the specific values of \(m\) and thus the equations of the tangents at point \(P\).