Circles
Tangent from External Point
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: A tangent line to a circle is perpendicular to the radius at the point of tangency. Use the condition that the distance from center O(0,0) to line AB equals the radius (3) to find the slope.
<p><strong>Step 1:</strong> Let the slope of AB be m. The line AB passes through A(-10, 0), so its equation is: y - 0 = m(x + 10), or mx - y + 10m = 0</p><p><strong>Step 2:</strong> The distance from center O(0, 0) to this line must equal the radius 3:</p><p>$$\frac{|m(0) - 0 + 10m|}{\sqrt{m^2 + 1}} = 3$$</p><p>$$\frac{|10m|}{\sqrt{m^2 + 1}} = 3$$</p><p><strong>Step 3:</strong> Square both sides: $$100m^2 = 9(m^2 + 1)$$</p><p>$$100m^2 = 9m^2 + 9$$</p><p>$$91m^2 = 9$$</p><p>$$m = \pm\frac{3}{\sqrt{91}}$$</p><p><strong>Step 4:</strong> Since B is in the third quadrant (x < 0, y < 0) and A is at (-10, 0), moving from A to B requires a negative slope. Both values work geometrically, but checking: with positive slope, B would be above the x-axis for x < -10, placing it outside third quadrant. With negative slope, B is below x-axis in the third quadrant.</p><p>∴ The slope of AB is $-\frac{3}{\sqrt{91}}$ or equivalently $-\frac{3\sqrt{91}}{91}$</p>
Correct Answer: A