Question

Let the circle C touch the line \(x - y + 1 = 0\), have the center on the positive x-axis, and cut off a chord of length \( \frac{4}{\sqrt{13}} \) along the line \( -3x + 2y = 1 \). Let H be the hyperbola \( \frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1 \), whose one of the foci is the center of C and the length of the transverse axis is the diameter of C. Then \( 2\alpha^2 + 3\beta^2 \) is equal to:

Hint:

To solve problems involving tangents, chords, and areas in circles and conic sections, set up geometric relationships and solve for unknowns using the relevant formulas for distance and areas.

Answer 1

Step 1: Information Analysis
The circle is tangent to the line \( x - y + 1 = 0 \).
The circle's center lies on the positive x-axis. The length of the chord formed by the line \( -3x + 2y = 1 \) is \( \frac{4}{\sqrt{13}} \).

Let the circle's center be \( C(\alpha, 0) \) and its radius be \( r \).

Step 2: Center-to-Line Distance Equation
The distance from \( C(\alpha, 0) \) to \( x - y + 1 = 0 \) is \( \frac{|\alpha - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|\alpha + 1|}{\sqrt{2}} \).
This distance equals the radius: \( \frac{|\alpha + 1|}{\sqrt{2}} = r \), which simplifies to \( (\alpha + 1)^2 = 2r^2 \) (Equation 1).

Step 3: Chord Length Equation
The chord length \( L \) on the line \( -3x + 2y = 1 \) is \( L = 2 \sqrt{r^2 - d^2} \), where \( d \) is the distance from the center to the line.

The distance \( d \) from \( C(\alpha, 0) \) to \( -3x + 2y - 1 = 0 \) is \( d = \frac{| -3\alpha + 2(0) - 1 |}{\sqrt{(-3)^2 + 2^2}} = \frac{| -3\alpha - 1 |}{\sqrt{13}} \).
Given \( L = \frac{4}{\sqrt{13}} \), we have \( \frac{4}{\sqrt{13}} = 2 \sqrt{r^2 - \left( \frac{| -3\alpha - 1 |}{\sqrt{13}} \right)^2} \).
Solving this equation yields a relationship between \( \alpha \) and \( r \).

Step 4: System Solution
Solving Equation 1 and the chord length equation simultaneously gives \( \alpha = \frac{-1}{5} \) and \( r = 2\sqrt{2} \).

Step 5: \( \alpha^2 \) and \( \beta^2 \) Calculation
Based on the derived relations, \( \alpha^2 = 8 \) and \( \beta^2 = 1 \).

Step 6: Final Computation
The expression \( 2\alpha^2 + 3\beta^2 \) evaluates to \( 2(8) + 3(1) = 16 + 3 = 19 \).