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

Given the equation of the tangent to \( y^2 = gx \) at the point \( (4, 6) \): 

The equation is: \( 3x - 4y + 12 = 0 \)

Also, the equation of the circle is: \( (x - 4)^2 + (y - 6)^2 + \lambda (3x - 4y + 12) = 0 \)

Expanding the equation of the circle:

\( x^2 + y^2 + (3x - 8)x + (-12 - 4\lambda)y + 52 + 12\lambda = 0 \)

Now, comparing the coefficients of \( x^2 \), \( y^2 \), and constant terms:

\( 2\sqrt{g^2} - c = 0 \Rightarrow g^2 = c \)

From the equation:

\( \left( \frac{3a - 8}{2} \right)^2 = 52 + 12\lambda \)

Now solve for \( \lambda \):

\( 9\lambda^2 + 64 - 48\lambda = 208 + 48\lambda \Rightarrow 9\lambda^2 - 96\lambda - 144 = 0 \)

Solving the quadratic equation:

\( \lambda = 12, -\frac{2}{3} \Rightarrow f = -30, -\frac{14}{3} \)

Now, the radius \( r \) is given by:

\( r = \sqrt{g^2 + f^2 - c} = |f| = |-(2\lambda + 6)| \)

Since the center lies in the second quadrant:

\( 3\lambda - 8 > 0 \Rightarrow \lambda > \frac{8}{3} \)

Thus, \( \lambda = 12, f = -30, r = 30 \)