<p>A circle \(C\) passes through the points of intersection of the parabola \(y + 1 = (x - 4)^2\) and the \(x\)-axis. The length of tangent from origin to \(C\) is:</p>
Step-by-Step Solution
Key Concept: Find intersection points of parabola with x-axis; construct circle through these points; use tangent length formula from external point.
<p>Intersection with \(x\)-axis: set \(y = 0\) in \(y + 1 = (x-4)^2\), giving \(1 = (x-4)^2\), so \(x = 3\) or \(x = 5\).</p><p>Points of intersection: \((3, 0)\) and \((5, 0)\).</p><p>Circle through \((3, 0)\) and \((5, 0)\) has center on perpendicular bisector at \(x = 4\). Let center be \((4, k)\).</p><p>Radius: \(r = \sqrt{(4-3)^2 + k^2} = \sqrt{1 + k^2}\)</p><p>For circle through these two points on \(x\)-axis, a general equation is \((x-3)(x-5) + y(y - 2k) = 0\).</p><p>Expanding: \(x^2 - 8x + 15 + y^2 - 2ky = 0\)</p><p>Length of tangent from origin: \(\sqrt{0^2 - 8(0) + 15 + 0^2 - 0} = \sqrt{15}\)</p>
Correct Answer: d