<p>If a chord of the circle \(x^2 + y^2 - 4x - 2y - c = 0\) is trisected at the points \((1/3,\, 1/3)\) and \((8/3,\, 8/3)\), then the radius of the circle will be:</p>
Step-by-Step Solution
Key Concept: If a chord is trisected at two points, the midpoint of those trisection points gives the midpoint of the chord. Use this with the perpendicularity condition (radius ⊥ chord at midpoint) to find the center, then calculate radius.
<p><strong>Step 1:</strong> Rewrite the circle equation in standard form: $(x-2)^2 + (y-1)^2 = c + 5$</p><p>Center is $O = (2, 1)$ and radius is $r = \sqrt{c+5}$</p><p><strong>Step 2:</strong> Let the trisection points be $P = (1/3, 1/3)$ and $Q = (8/3, 8/3)$. The midpoint $M$ of $PQ$ is: $M = \left(\frac{1/3 + 8/3}{2}, \frac{1/3 + 8/3}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}\right)$</p><p><strong>Step 3:</strong> This point $M$ is the midpoint of the chord. The radius from center $O(2,1)$ to $M$ must be perpendicular to the chord $PQ$.</p><p>Slope of $PQ = \frac{8/3 - 1/3}{8/3 - 1/3} = 1$</p><p>Slope of $OM = \frac{3/2 - 1}{3/2 - 2} = \frac{1/2}{-1/2} = -1$ ✓ (perpendicular confirmed)</p><p><strong>Step 4:</strong> Since both $P$ and $Q$ lie on the circle: $(1/3)^2 + (1/3)^2 - 4(1/3) - 2(1/3) - c = 0$</p><p>$\frac{2}{9} - \frac{6}{3} - c = 0 \Rightarrow \frac{2}{9} - 2 - c = 0 \Rightarrow c = -\frac{16}{9}$</p><p><strong>Step 5:</strong> Therefore $r = \sqrt{c + 5} = \sqrt{-\frac{16}{9} + 5} = \sqrt{\frac{29}{9}} = \frac{\sqrt{29}}{3}$</p><p>∴ Answer: C</p>
Correct Answer: C