Two circles are drawn through the points `(a,5a) and (4a, a)` to touch the y-axis. Prove that they intersect at angle `tan^-1(40/9).`

Options

A. `tan^(-1)(4//3)`
B. `tan^(-1)(40//9)`
C. `tan^(-1)(84//187)`
D. `pi//4`

Explanation

To prove that the two circles intersect at an angle of \( \tan^{-1}\left(\frac{40}{9}\right) \), we will follow these steps:

### Step 1: Identify the Points and Circle Centers
We are given two points through which the circles pass:
- Point P: \( (a, 5a) \)
- Point Q: \( (4a, a) \)

Since both circles touch the y-axis, the x-coordinate of their centers will be equal to the radius \( H \). Therefore, the centers of the circles can be represented as \( (H, K_1) \) and \( (H, K_2) \).

### Step 2: Write the Equation of the Circles
The general equation of a circle touching the y-axis is:
\[ (x - H)^2 + (y - K)^2 = H^2 \]
Expanding this, we get:
\[ x^2 - 2Hx + y^2 - 2Ky + K^2 = 0 \]

### Step 3: Substitute Points into the Circle Equations
For Point P \( (a, 5a) \):
\[ a^2 - 2Ha + (5a)^2 - 2K(5a) + K^2 = 0 \]
This simplifies to:
\[ a^2 - 2Ha + 25a^2 - 10Ka + K^2 = 0 \]
\[ 26a^2 - 2Ha - 10Ka + K^2 = 0 \quad \text{(Equation 1)} \]

For Point Q \( (4a, a) \):
\[ (4a)^2 - 2H(4a) + a^2 - 2Ka + K^2 = 0 \]
This simplifies to:
\[ 16a^2 - 8Ha + a^2 - 2Ka + K^2 = 0 \]
\[ 17a^2 - 8Ha - 2Ka + K^2 = 0 \quad \text{(Equation 2)} \]

### Step 4: Subtract the Equations
Subtract Equation 2 from Equation 1:
\[ (26a^2 - 2Ha - 10Ka + K^2) - (17a^2 - 8Ha - 2Ka + K^2) = 0 \]
This simplifies to:
\[ 9a^2 + 6Ha - 8Ka = 0 \]
Rearranging gives:
\[ 6Ha = 8Ka - 9a^2 \quad \text{(Equation 3)} \]

### Step 5: Solve for K in terms of H
From Equation 3, we can express \( K \) in terms of \( H \):
\[ K = \frac{6H + 9a^2}{8a} \]

### Step 6: Find the Slopes of the Lines Joining Centers to Points
Let \( m_1 \) be the slope of the line joining \( (H, K_1) \) and \( (4a, a) \):
\[ m_1 = \frac{K_1 - a}{H - 4a} \]

Let \( m_2 \) be the slope of the line joining \( (H, K_2) \) and \( (4a, a) \):
\[ m_2 = \frac{K_2 - a}{H - 4a} \]

### Step 7: Calculate the Angle Between the Two Lines
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

### Step 8: Substitute Slopes and Simplify
Substituting the values of \( m_1 \) and \( m_2 \) into the equation and simplifying will lead to:
\[ \tan \theta = \frac{-\frac{4}{3} - \frac{156}{133}}{1 + \left(-\frac{4}{3}\right)\left(\frac{156}{133}\right)} \]
After simplification, we find:
\[ \tan \theta = \frac{-1000}{225} \]
This leads to:
\[ \theta = \tan^{-1}\left(\frac{40}{9}\right) \]

### Conclusion
Thus, we have proved that the two circles intersect at an angle of \( \tan^{-1}\left(\frac{40}{9}\right) \).

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