Circles
Locus of midpoint
Grade None
Question:
<p>If a tangent to the circle \(x^2 + y^2 = 1\) intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is:</p>
<p>\(x^2 + y^2 - 4x^2y^2 = 0\)</p>
<p>\(x^2 + y^2 - 2xy = 0\)</p>
<p>\(x^2 + y^2 - 16x^2y^2 = 0\)</p>
<p>\(x^2 + y^2 - 2x^2y^2 = 0\)</p>
Step-by-Step Solution
Key Concept: A tangent to the unit circle has the form xa + yb = 1 (where a² + b² = 1), which directly gives intercepts at P(a, 0) and Q(0, b). The midpoint traces a path determined by the constraint a² + b² = 1.
<p><strong>Step 1:</strong> Write the tangent to circle x² + y² = 1 in intercept form. If tangent meets x-axis at P(a, 0) and y-axis at Q(0, b), the tangent equation is: <strong>x/a + y/b = 1</strong></p><p><strong>Step 2:</strong> Since this line is tangent to x² + y² = 1, use the condition that distance from origin to line equals radius (1):<br>Distance = |1|/√(1/a² + 1/b²) = 1<br>This gives: <strong>1/a² + 1/b² = 1</strong></p><p><strong>Step 3:</strong> Let the midpoint of PQ be M(h, k), so:<br>h = a/2 ⟹ a = 2h<br>k = b/2 ⟹ b = 2k</p><p><strong>Step 4:</strong> Substitute into the constraint 1/a² + 1/b² = 1:<br>1/(4h²) + 1/(4k²) = 1<br><strong>1/h² + 1/k² = 4</strong><br>Or equivalently: <strong>x² + y² = (1/4)(x²y²)/(x² + y²)</strong></p><p>In standard form: <strong>1/x² + 1/y² = 4</strong></p><p>∴ Answer: D</p>
Correct Answer: D