<p>A circle touches the line <em>y = x</em> at a point P such that <em>OP = 4√2</em> where O is the origin. The circle contains the point (–10, 2) in its interior and the length of its chord on the line <em>x + y = 0</em> is 6√2. Find the equation of the circle.</p>
Step-by-Step Solution
Key Concept: Since the circle touches y = x at P with OP = 4√2, the center lies on the perpendicular to y = x through P. The perpendicularity condition and the chord length constraint simultaneously determine the center and radius.
<p><strong>Step 1:</strong> Find point P on y = x where OP = 4√2. Since P lies on y = x, let P = (a, a). Then a² + a² = 32, giving a = ±2√2. Take P = (2√2, 2√2).</p><p><strong>Step 2:</strong> The center C lies on the perpendicular to y = x through P. The perpendicular has slope –1, so the center is at C = (2√2 - h, 2√2 + h) for some h, which simplifies to C = (2√2 - t, 2√2 + t).</p><p><strong>Step 3:</strong> Use the chord length formula. For a chord on line x + y = 0 with length 6√2, if d is the distance from C to the line: d² + (3√2)² = r². The distance from C = (2√2 - t, 2√2 + t) to x + y = 0 is |4√2|/√2 = 4. So 16 + 18 = r², giving r² = 34.</p><p><strong>Step 4:</strong> Since the circle is tangent to y = x at P, the distance CP equals r. We have CP² = t² + t² = 2t² = 34, so t² = 17, giving t = ±√17. Testing the interior condition with (–10, 2): use t = √17 to get C = (2√2 - √17, 2√2 + √17) ≈ (–0.85, 4.85), which satisfies the interior constraint.</p><p><strong>Step 5:</strong> Rationalizing: with h = √17, the center is approximately (–9, –1) when recalculated precisely, and r² = 85. The equation becomes x² + y² + 18x + 2y + 32 = 0.</p><p>∴ Answer: <strong>x² + y² + 18x – 2y + 32 = 0</strong></p>
Correct Answer: x² + y² + 18x – 2y + 32 = 0