<p>A circle touching the X-axis at (3, 0) and making an intercept of length 8 on the Y-axis passes through the point</p>
Step-by-Step Solution
Key Concept: When a circle touches the X-axis, its centre lies on a vertical line through the point of tangency. Use the intercept property to find the radius.
<p><strong>Step 1:</strong> Since the circle touches the X-axis at (3, 0), the centre lies on the vertical line x = 3. Let the centre be (3, r) where r is the radius.</p><p><strong>Step 2:</strong> The circle makes an intercept of 8 on the Y-axis. If M is the midpoint of this intercept on the Y-axis, then AM = BM = 4, and CM = 3 (horizontal distance from centre to Y-axis).</p><p><strong>Step 3:</strong> Using the relationship for a chord: \(r^2 = CM^2 + AM^2 = 9 + 16 = 25\), so \(r = 5\)</p><p><strong>Step 4:</strong> The equation of the circle is \((x - 3)^2 + (y - 5)^2 = 25\)</p><p><strong>Step 5:</strong> Checking option (a): (3, 10): \((3-3)^2 + (10-5)^2 = 0 + 25 = 25\) ✓</p><p>∴ Answer is (a) (3, 10)</p>
Correct Answer: A