To find the centers of the two circles that pass through the points \( P(a, 5a) \) and \( Q(4a, a) \) and touch the y-axis, we can follow these steps:
### Step 1: Define the center of the circles
Let the center of the circles be \( (H, K) \). Since the circles touch the y-axis, the distance from the center to the y-axis is equal to the x-coordinate of the center, which is \( H \). Therefore, the equation of the circle can be expressed as:
\[
(x - H)^2 + (y - K)^2 = H^2
\]
### Step 2: Substitute the points into the circle equation
Since the circle passes through point \( P(a, 5a) \), we substitute \( x = a \) and \( y = 5a \) into the circle equation:
\[
(a - H)^2 + (5a - K)^2 = H^2
\]
Expanding this gives:
\[
(a^2 - 2aH + H^2) + (25a^2 - 10aK + K^2) = H^2
\]
Simplifying, we get:
\[
26a^2 - 2aH - 10aK + K^2 = 0 \quad \text{(Equation 1)}
\]
### Step 3: Substitute the second point into the circle equation
Next, we substitute point \( Q(4a, a) \) into the circle equation:
\[
(4a - H)^2 + (a - K)^2 = H^2
\]
Expanding this gives:
\[
(16a^2 - 8aH + H^2) + (a^2 - 2aK + K^2) = H^2
\]
Simplifying, we get:
\[
17a^2 - 8aH - 2aK + K^2 = 0 \quad \text{(Equation 2)}
\]
### Step 4: Subtract Equation 2 from Equation 1
Now, we subtract Equation 2 from Equation 1:
\[
(26a^2 - 2aH - 10aK + K^2) - (17a^2 - 8aH - 2aK + K^2) = 0
\]
This simplifies to:
\[
9a^2 + 6aH - 8aK = 0
\]
Rearranging gives:
\[
9a^2 = 8aK - 6aH
\]
Dividing through by \( a \) (assuming \( a \neq 0 \)):
\[
9a = 8K - 6H \quad \text{(Equation 3)}
\]
### Step 5: Express \( K \) in terms of \( H \)
From Equation 3, we can express \( K \):
\[
8K = 9a + 6H \implies K = \frac{9a + 6H}{8}
\]
### Step 6: Substitute \( K \) back into one of the original equations
Substituting \( K \) back into Equation 1:
\[
26a^2 - 2aH - 10\left(\frac{9a + 6H}{8}\right) + \left(\frac{9a + 6H}{8}\right)^2 = 0
\]
This will lead to a quadratic equation in terms of \( H \).
### Step 7: Solve for \( H \)
After simplifying the above equation, we will find two possible values for \( H \).
### Step 8: Find corresponding \( K \) values
Using the values of \( H \) found, substitute back into the expression for \( K \) to find the corresponding \( K \) values.
### Final Centers
The centers of the circles will be:
1. \( \left( \frac{5a}{2}, 3a \right) \)
2. \( \left( \frac{205a}{18}, \frac{29a}{3} \right) \)
