Circles
Common Tangents
Grade 11
Question:
<p>If <span class="math">\(a > 2b > 0\)</span> then the positive value of m for which <span class="math">\(y = mx - b\sqrt{1 + m^2}\)</span> is a common tangent to <span class="math">\(x^2 + y^2 = b^2\)</span> and <span class="math">\((x - a)^2 + y^2 = b^2\)</span> is:</p>
<p>(a) \(\frac{2b}{\sqrt{a^2 - 4b^2}}\)</p>
<p>(b) \(\frac{a^2 - 4b^2}{2b}\)</p>
<p>(c) \(\frac{2b}{a - 2b}\)</p>
<p>(d) \(\frac{b}{a - 2b}\)</p>
Step-by-Step Solution
Key Concept: A line is tangent to a circle if the perpendicular distance from the circle's center to the line equals the radius. Since the given line is a common tangent to both circles, this distance condition must hold for both circles simultaneously.
<p><strong>Step 1: Identify the circles and tangent line.</strong></p><p>Circle 1: x² + y² = b² (center C₁ = (0,0), radius r₁ = b)</p><p>Circle 2: (x-a)² + y² = b² (center C₂ = (a,0), radius r₂ = b)</p><p>Tangent line: y = mx - b√(1+m²), or mx - y - b√(1+m²) = 0</p><p><strong>Step 2: Apply the distance condition for Circle 1.</strong></p><p>Distance from (0,0) to line mx - y - b√(1+m²) = 0:</p><p>d₁ = |m(0) - 0 - b√(1+m²)|/√(m²+1) = b√(1+m²)/√(m²+1) = b</p><p>This equals r₁ = b ✓ (tangent condition is satisfied for Circle 1)</p><p><strong>Step 3: Apply the distance condition for Circle 2.</strong></p><p>Distance from (a,0) to line mx - y - b√(1+m²) = 0:</p><p>d₂ = |m(a) - 0 - b√(1+m²)|/√(m²+1) = |ma - b√(1+m²)|/√(m²+1)</p><p>For tangency: d₂ = b</p><p>So: |ma - b√(1+m²)|/√(m²+1) = b</p><p><strong>Step 4: Solve for m.</strong></p><p>|ma - b√(1+m²)| = b√(m²+1)</p><p>Since m > 0 and a > 2b > 0, we have ma > b√(1+m²) (as the line must be external tangent)</p><p>Therefore: ma - b√(1+m²) = b√(m²+1)</p><p>ma = b√(1+m²) + b√(m²+1) = 2b√(1+m²)</p><p>Squaring both sides:</p><p>m²a² = 4b²(1+m²)</p><p>m²a² = 4b² + 4b²m²</p><p>m²(a² - 4b²) = 4b²</p><p>m² = 4b²/(a² - 4b²)</p><p>m = 2b/√(a² - 4b²) (taking positive value)</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a