To solve the problem, we need to find the coordinates of the point of intersection of the tangents drawn at points P and Q, where the line intersects the circle. Let's break this down step by step.
### Step 1: Find the points of intersection (P and Q)
We have the equations:
1. Circle: \( x^2 + y^2 = 25 \)
2. Line: \( x - 2y + 1 = 0 \)
First, we can express \( x \) in terms of \( y \) from the line equation:
\[
x = 2y - 1
\]
Now, substitute this expression for \( x \) into the circle equation:
\[
(2y - 1)^2 + y^2 = 25
\]
Expanding this:
\[
4y^2 - 4y + 1 + y^2 = 25
\]
\[
5y^2 - 4y + 1 - 25 = 0
\]
\[
5y^2 - 4y - 24 = 0
\]
### Step 2: Solve the quadratic equation for \( y \)
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5 \), \( b = -4 \), and \( c = -24 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-4)^2 - 4 \cdot 5 \cdot (-24) = 16 + 480 = 496
\]
Now substituting into the quadratic formula:
\[
y = \frac{4 \pm \sqrt{496}}{10}
\]
\[
y = \frac{4 \pm 4\sqrt{31}}{10} = \frac{2 \pm 2\sqrt{31}}{5}
\]
### Step 3: Find corresponding \( x \) values
Using \( y = \frac{2 + 2\sqrt{31}}{5} \) and \( y = \frac{2 - 2\sqrt{31}}{5} \), we can find \( x \):
For \( y_1 = \frac{2 + 2\sqrt{31}}{5} \):
\[
x_1 = 2y_1 - 1 = 2\left(\frac{2 + 2\sqrt{31}}{5}\right) - 1 = \frac{4 + 4\sqrt{31}}{5} - 1 = \frac{4 + 4\sqrt{31} - 5}{5} = \frac{-1 + 4\sqrt{31}}{5}
\]
For \( y_2 = \frac{2 - 2\sqrt{31}}{5} \):
\[
x_2 = 2y_2 - 1 = 2\left(\frac{2 - 2\sqrt{31}}{5}\right) - 1 = \frac{4 - 4\sqrt{31}}{5} - 1 = \frac{4 - 4\sqrt{31} - 5}{5} = \frac{-1 - 4\sqrt{31}}{5}
\]
Thus, the points of intersection \( P \) and \( Q \) are:
\[
P\left(\frac{-1 + 4\sqrt{31}}{5}, \frac{2 + 2\sqrt{31}}{5}\right), \quad Q\left(\frac{-1 - 4\sqrt{31}}{5}, \frac{2 - 2\sqrt{31}}{5}\right)
\]
### Step 4: Find the coordinates of the point of intersection of the tangents at P and Q
The coordinates of the point of intersection of the tangents drawn at points \( P \) and \( Q \) can be found using the formula:
\[
\left( \frac{h}{25}, \frac{k}{25} \right)
\]
where \( h \) and \( k \) are the coordinates of the center of the circle, which is \( (0, 0) \) in this case.
Thus, the coordinates of the intersection of the tangents is:
\[
\left( \frac{0}{25}, \frac{0}{25} \right) = (0, 0)
\]
### Final Answer:
The coordinates of the point of intersection of the tangents drawn at points P and Q to the circle is \( (0, 0) \).
