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

We have two equations: 
\((4-a)^2 + (6-r)^2 = r^2\)  
\(4a + 3r = 34\) 
From equation (2), \(a = \frac{34 - 3r}{4}\).
Substituting this into equation (1): \[ \left(4 - \frac{34-3r}{4}\right)^2 + (6-r)^2 = r^2 \] \[ \left(\frac{16 - 34 + 3r}{4}\right)^2 + (6-r)^2 = r^2 \] \[ \left(\frac{-18 + 3r}{4}\right)^2 + (6-r)^2 = r^2 \] \[ \frac{9}{16}(r-6)^2 + (r-6)^2 = r^2 \] \[ \frac{9}{16}(r^2 - 12r + 36) + r^2 - 12r + 36 = r^2 \] \[ \frac{9}{16}r^2 - \frac{27}{4}r + \frac{81}{4} + r^2 - 12r + 36 = r^2 \] \[ \frac{9}{16}r^2 + r^2 - r^2 - \frac{27}{4}r - 12r + \frac{81}{4} + 36 = 0 \] \[ \frac{9}{16}r^2 - \left(\frac{27}{4} + \frac{48}{4}\right)r + \left(\frac{81}{4} + \frac{144}{4}\right) = 0 \] \[ \frac{9}{16}r^2 - \frac{75}{4}r + \frac{225}{4} = 0 \] Multiply by \(\frac{16}{9}\): \[ r^2 - \frac{75}{4} \cdot \frac{16}{9}r + \frac{225}{4} \cdot \frac{16}{9} = 0 \] \[ r^2 - \frac{300}{9}r + \frac{3600}{36} = 0 \] \[ r^2 - \frac{100}{3}r + 100 = 0 \] \[ 3r^2 - 100r + 300 = 0 \] Now use the quadratic formula: \[ r = \frac{100 \pm \sqrt{10000 - 4 \cdot 3 \cdot 300}}{2 \cdot 3} \] \[ r = \frac{100 \pm \sqrt{10000 - 3600}}{6} \] \[ r = \frac{100 \pm \sqrt{6400}}{6} \] \[ r = \frac{100 \pm 80}{6} \] So, \(r = \frac{100 + 80}{6} = \frac{180}{6} = 30\) or \(r = \frac{100 - 80}{6} = \frac{20}{6} = \frac{10}{3}\).
If \(r = 30\), then \(a = \frac{34 - 3 \cdot 30}{4} = \frac{34 - 90}{4} = \frac{-56}{4} = -14\).
If \(r = \frac{10}{3}\), then \(a = \frac{34 - 3 \cdot \frac{10}{3}}{4} = \frac{34 - 10}{4} = \frac{24}{4} = 6\).
But \(a\) must be negative.
So \(r = 30\) and \(a = -14\).
Therefore, \(r = 30\).