Circles
Tangent to circle
Grade 11
Question:
<p>If the tangent to the conic, \(y - 6 = x^2\) at (2, 10) touches the circle, \(x^2 + y^2 + 8x - 2y = k\) (for some fixed \(k\)) at a point \((\alpha, \beta)\), then \((\alpha, \beta)\) is</p>
<p>\(\left(-\dfrac{6}{17}, \dfrac{10}{17}\right)\)</p>
<p>\(\left(-\dfrac{8}{17}, \dfrac{2}{17}\right)\)</p>
<p>\(\left(-\dfrac{4}{17}, \dfrac{1}{17}\right)\)</p>
<p>\(\left(-\dfrac{7}{17}, \dfrac{6}{17}\right)\)</p>
Step-by-Step Solution
Key Concept: The tangent line to the parabola at a point is also tangent to the circle, meaning it touches the circle at exactly one point. The radius to that point of tangency must be perpendicular to the tangent line.
<p><strong>Step 1:</strong> Find the tangent to the parabola y - 6 = x² at (2, 10).</p><p>Differentiating: dy/dx = 2x. At x = 2: dy/dx = 4.</p><p>Tangent line equation: y - 10 = 4(x - 2) ⟹ <strong>4x - y + 2 = 0</strong></p><p><strong>Step 2:</strong> Rewrite the circle equation in standard form.</p><p>x² + y² + 8x - 2y = k ⟹ (x + 4)² + (y - 1)² = k + 17</p><p>Center: C = (-4, 1), Radius: r = √(k + 17)</p><p><strong>Step 3:</strong> The point of tangency (α, β) lies on the radius perpendicular to the tangent line 4x - y + 2 = 0.</p><p>The radius has slope -1/4 (perpendicular to slope 4 of tangent).</p><p>Radius line through C(-4, 1): y - 1 = -1/4(x + 4)</p><p><strong>Step 4:</strong> Find intersection of radius with tangent line.</p><p>From tangent: y = 4x + 2. Substitute into radius equation:</p><p>4x + 2 - 1 = -1/4(x + 4) ⟹ 4x + 1 = -x/4 - 1 ⟹ 17x/4 = -2 ⟹ x = -8/17</p><p>y = 4(-8/17) + 2 = -32/17 + 34/17 = 2/17</p><p>∴ Answer: (α, β) = (-8/17, 2/17) which is <strong>D</strong></p>
Correct Answer: D