Circles
Tangent from external point to circle
Grade None
Question:
<p>Find the slope(s) of the tangent(s) drawn from the origin to the circle \((x-3)^2 + (y-3)^2 = 6\).</p>
<p>A. \(m = 3 \pm \sqrt{2}\)</p>
<p>B. \(m = 3 \pm 2\sqrt{2}\)</p>
<p>C. \(m = 1 \pm \sqrt{2}\)</p>
<p>D. \(m = 2 \pm \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: A tangent from external point to circle is perpendicular to the radius at point of contact. Use the condition that distance from center to tangent line equals the radius to find slope.
<p><strong>Step 1:</strong> Let tangent from origin O(0,0) have slope m, so equation is y = mx or mx - y = 0.</p><p><strong>Step 2:</strong> Distance from center C(3,3) to line mx - y = 0 must equal radius √6:</p><p>$$\frac{|3m - 3|}{\sqrt{m^2 + 1}} = \sqrt{6}$$</p><p><strong>Step 3:</strong> Square both sides:</p><p>$$\frac{(3m-3)^2}{m^2+1} = 6$$</p><p>$$9(m-1)^2 = 6(m^2+1)$$</p><p>$$9m^2 - 18m + 9 = 6m^2 + 6$$</p><p>$$3m^2 - 18m + 3 = 0$$</p><p>$$m^2 - 6m + 1 = 0$$</p><p><strong>Step 4:</strong> Using quadratic formula:</p><p>$$m = \frac{6 \pm \sqrt{36-4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}$$</p><p>∴ Answer: B</p>
Correct Answer: B