To solve the problem step by step, we will follow the outlined approach in the video transcript while providing clear explanations for each step.
### Step 1: Find the equation of the tangent to Circle C1 at point (1, 2)
The equation of the circle \( C_1 \) is given by:
\[
x^2 + y^2 = 5
\]
The general equation of the tangent to a circle \( x^2 + y^2 = r^2 \) at point \( (x_1, y_1) \) is given by:
\[
xx_1 + yy_1 = r^2
\]
For our circle \( C_1 \), \( r^2 = 5 \) and the point is \( (1, 2) \). Substituting these values into the tangent equation:
\[
x \cdot 1 + y \cdot 2 = 5
\]
This simplifies to:
\[
x + 2y = 5
\]
### Step 2: Find the points of intersection A and B with Circle C2
The equation of the second circle \( C_2 \) is:
\[
x^2 + y^2 = 9
\]
We will substitute the equation of the tangent \( x + 2y = 5 \) into the equation of the circle \( C_2 \). First, solve for \( x \):
\[
x = 5 - 2y
\]
Now substitute \( x \) into the equation of \( C_2 \):
\[
(5 - 2y)^2 + y^2 = 9
\]
Expanding this:
\[
25 - 20y + 4y^2 + y^2 = 9
\]
Combining like terms:
\[
5y^2 - 20y + 25 - 9 = 0
\]
This simplifies to:
\[
5y^2 - 20y + 16 = 0
\]
### Step 3: Solve the quadratic equation for y
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5, b = -20, c = 16 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-20)^2 - 4 \cdot 5 \cdot 16 = 400 - 320 = 80
\]
Now substituting into the quadratic formula:
\[
y = \frac{20 \pm \sqrt{80}}{10} = \frac{20 \pm 4\sqrt{5}}{10} = 2 \pm \frac{2\sqrt{5}}{5}
\]
Thus, we have two values for \( y \):
\[
y_1 = 2 + \frac{2\sqrt{5}}{5}, \quad y_2 = 2 - \frac{2\sqrt{5}}{5}
\]
### Step 4: Find corresponding x-coordinates for points A and B
Substituting \( y_1 \) and \( y_2 \) back into \( x = 5 - 2y \):
For \( y_1 \):
\[
x_1 = 5 - 2\left(2 + \frac{2\sqrt{5}}{5}\right) = 5 - 4 - \frac{4\sqrt{5}}{5} = 1 - \frac{4\sqrt{5}}{5}
\]
For \( y_2 \):
\[
x_2 = 5 - 2\left(2 - \frac{2\sqrt{5}}{5}\right) = 5 - 4 + \frac{4\sqrt{5}}{5} = 1 + \frac{4\sqrt{5}}{5}
\]
Thus, the points A and B are:
\[
A\left(1 - \frac{4\sqrt{5}}{5}, 2 + \frac{2\sqrt{5}}{5}\right), \quad B\left(1 + \frac{4\sqrt{5}}{5}, 2 - \frac{2\sqrt{5}}{5}\right)
\]
### Step 5: Find the equation of the tangents at points A and B
The general equation of the tangent to circle \( C_2 \) at point \( (x_1, y_1) \) is:
\[
xx_1 + yy_1 = r^2
\]
For points A and B, we can derive the equations of the tangents. However, we will focus on finding the intersection point C of these tangents.
### Step 6: Find the coordinates of point C
From the tangents at points A and B, we can derive that the coordinates of point C can be found through the intersection of the tangents.
Using the relationship derived from the tangents, we can find:
\[
\frac{h}{1} = \frac{k}{2} = \frac{9}{5}
\]
Solving this gives:
\[
h = \frac{9}{5}, \quad k = \frac{18}{5}
\]
Thus, the coordinates of point C are:
\[
C\left(\frac{9}{5}, \frac{18}{5}\right)
\]
### Final Answer
The coordinates of point C are:
\[
\left(\frac{9}{5}, \frac{18}{5}\right)
\]
