Question

Let $ r $ be the radius of the circle, which touches the $ x $-axis at point $ (a, 0) $, $ a < 0 $ and the parabola $ y^2 = 9x $ at the point $ (4, 6) $. Then $ r $ is equal to:

Hint:

When solving tangency problems involving a circle and a parabola, ensure the point of tangency satisfies both the equation of the circle and the parabola, and use the geometric relationship between the circle and the tangent line.

Answer 1

Correct Answer: 30

The equation of the tangent to \( y^2 = gx \) at the point \( (4, 6) \) is given as \( 3x - 4y + 12 = 0 \).

The equation of a circle is provided as \( (x - 4)^2 + (y - 6)^2 + \lambda (3x - 4y + 12) = 0 \).

Expanding the circle's equation yields: \( x^2 + y^2 + (3\lambda - 8)x + (-4\lambda - 12)y + 52 + 12\lambda = 0 \).

Comparing coefficients of \( x^2 \), \( y^2 \), and the constant term, we use the relationship \( 2\sqrt{g^2} - c = 0 \), which simplifies to \( g^2 = c \).

From the expanded equation, we derive: \( \left( \frac{3\lambda - 8}{2} \right)^2 = 52 + 12\lambda \).

Solving for \( \lambda \) leads to the quadratic equation \( 9\lambda^2 - 96\lambda - 144 = 0 \).

The solutions for \( \lambda \) are \( \lambda = 12 \) and \( \lambda = -\frac{2}{3} \). Consequently, \( f = -(2\lambda + 6) \) yields \( f = -30 \) and \( f = -\frac{14}{3} \).

The radius \( r \) is calculated as \( r = \sqrt{g^2 + f^2 - c} = |f| = |-(2\lambda + 6)| \).

Given that the center of the circle lies in the second quadrant, the condition \( 3\lambda - 8>0 \) implies \( \lambda>\frac{8}{3} \).

Therefore, the valid solution is \( \lambda = 12 \), which results in \( f = -30 \) and a radius \( r = 30 \).