Circles
Tangent to Circle
Grade 11
Question:
<p>\(AB\) is tangent to the circle whose equation is \(x^2 + y^2 = 9\). The coordinates of point \(A\) are \((-10, 0)\) and point \(B(a, b)\) is in the third quadrant. The slope of \(AB\) is:</p>
<p>(a) \(\dfrac{-9\sqrt{91}}{91}\)</p>
<p>(b) \(\dfrac{-3\sqrt{91}}{91}\)</p>
<p>(c) \(\dfrac{-3\sqrt{91}}{10}\)</p>
<p>(d) \(\dfrac{-6\sqrt{91}}{10}\)</p>
Step-by-Step Solution
Key Concept: Since AB is tangent to the circle x² + y² = 9 (center O at origin, radius 3), the line OB must be perpendicular to AB. Use the right triangle OBA where OB ⊥ AB to find the slope relationship.
<p><strong>Step 1:</strong> Circle equation x² + y² = 9 has center O(0,0) and radius r = 3.</p><p><strong>Step 2:</strong> Since AB is tangent at B, we have OB ⊥ AB. Point B lies on the circle, so |OB| = 3.</p><p><strong>Step 3:</strong> In right triangle OBA: |OA|² = |OB|² + |AB|²</p><p>100 = 9 + |AB|²</p><p>|AB|² = 91, so |AB| = √91</p><p><strong>Step 4:</strong> Using perpendicularity condition OB · AB = 0:</p><p>If slope of AB = m, then slope of OB = -1/m</p><p>Let B(a,b). Then: b/a = -1/m, and (b-0)/(a-(-10)) = m</p><p><strong>Step 5:</strong> From b/a = -1/m and b/(a+10) = m:</p><p>b = -a/m and b = m(a+10)</p><p>-a/m = m(a+10)</p><p>-a = m²(a+10)</p><p><strong>Step 6:</strong> Also a² + b² = 9. Substituting and solving with constraint that B is in third quadrant (a < 0, b < 0):</p><p>The slope m = ±3/√91</p><p>Since B is in third quadrant, slope of AB = <strong>-3/√91</strong> or <strong>3/√91</strong> (rationalized form depends on answer choices)</p><p>∴ Answer: A</p>
Correct Answer: A