Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P,Q} is equal to

Options

A. `{(4,0),(0,6)}`
B. `{(2+2 sqrt2, 3-sqrt5), (2-2 sqrt2, 3+sqrt5)}`
C. `{(2 +2 sqrt2, 3+sqrt5), (2-2 sqrt2, 3-sqrt5)}`
D. `{(-1, 5), (5,1)}`

Answer

To solve the problem, we need to find the coordinates of points P and Q on a circle centered at C(2, 3) that passes through the origin O(0, 0). The line segment OC is perpendicular to both CP and CQ.

### Step-by-Step Solution:

1. **Find the Radius of the Circle**:
The radius \( r \) of the circle can be found using the distance formula between the center C(2, 3) and the origin O(0, 0):
\[ r = \sqrt{(2 - 0)^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

2. **Write the Equation of the Circle**:
The standard form of the equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \( h = 2 \), \( k = 3 \), and \( r^2 = 13 \). Thus, the equation becomes:
\[ (x - 2)^2 + (y - 3)^2 = 13 \]
Expanding this, we get:
\[ x^2 - 4x + 4 + y^2 - 6y + 9 = 13 \]
Simplifying, we have:
\[ x^2 + y^2 - 4x - 6y = 0 \quad \text{(Equation 1)} \]

3. **Find the Slope of OC**:
The slope of line segment OC is calculated as:
\[ \text{slope of } OC = \frac{3 - 0}{2 - 0} = \frac{3}{2} \]

4. **Find the Slope of Line PQ**:
Since OC is perpendicular to both CP and CQ, the slope of line PQ is the negative reciprocal of the slope of OC:
\[ \text{slope of } PQ = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \]

5. **Write the Equation of Line PQ**:
Using the point-slope form of the line equation, where the point is C(2, 3):
\[ y - 3 = -\frac{2}{3}(x - 2) \]
Rearranging gives:
\[ y - 3 = -\frac{2}{3}x + \frac{4}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y - 9 = -2x + 4 \]
Rearranging gives:
\[ 2x + 3y = 13 \quad \text{(Equation 2)} \]

6. **Solve the System of Equations**:
We now have two equations:
- Equation 1: \( x^2 + y^2 - 4x - 6y = 0 \)
- Equation 2: \( 2x + 3y = 13 \)

We can express \( y \) from Equation 2:
\[ y = \frac{13 - 2x}{3} \]
Substitute this into Equation 1:
\[ x^2 + \left(\frac{13 - 2x}{3}\right)^2 - 4x - 6\left(\frac{13 - 2x}{3}\right) = 0 \]
Expanding and simplifying leads to a quadratic equation in \( x \).

7. **Finding the Roots**:
After solving the quadratic equation, we find the values of \( x \):
\[ x = 5 \quad \text{and} \quad x = -1 \]

8. **Finding Corresponding y-values**:
Substitute \( x = 5 \) into Equation 2:
\[ 2(5) + 3y = 13 \implies 10 + 3y = 13 \implies 3y = 3 \implies y = 1 \]
So one point is \( P(5, 1) \).

Substitute \( x = -1 \):
\[ 2(-1) + 3y = 13 \implies -2 + 3y = 13 \implies 3y = 15 \implies y = 5 \]
So the other point is \( Q(-1, 5) \).

### Final Answer:
The set of points \( \{P, Q\} \) is:
\[ \{(5, 1), (-1, 5)\} \]