<p>A circle touching the <em>x</em>-axis at (3, 0) and making an intercept of length 8 on the <em>y</em>-axis passes through the point</p>
Step-by-Step Solution
Key Concept: If a circle touches the x-axis at (3, 0), its center must be at (3, r) where r is the radius. Use the intercept condition on the y-axis to find r, then verify which point satisfies the circle equation.
<p><strong>Step 1:</strong> Since the circle touches the x-axis at (3, 0), the center is at C(3, r) where r is the radius.</p><p><strong>Step 2:</strong> The circle equation is (x - 3)² + (y - r)² = r².</p><p><strong>Step 3:</strong> For the y-axis intercept, substitute x = 0: (0 - 3)² + (y - r)² = r², which gives 9 + (y - r)² = r², so (y - r)² = r² - 9, yielding y = r ± √(r² - 9).</p><p><strong>Step 4:</strong> The intercept length is |y₁ - y₂| = 2√(r² - 9) = 8, so √(r² - 9) = 4, giving r² - 9 = 16, thus r² = 25 and r = 5.</p><p><strong>Step 5:</strong> Circle equation: (x - 3)² + (y - 5)² = 25. The y-intercepts are at y = 5 ± 4, i.e., y = 9 and y = 1.</p><p><strong>Step 6:</strong> Test the given options by substituting into (x - 3)² + (y - 5)² = 25 to find which point lies on the circle.</p><p>∴ Answer: A</p>
Correct Answer: A