Circles
Tangent to Circle
Grade 11
Question:
<p>We have two straight lines \(x - y = 1\) and \(2x + y = 3\). The tangent to the circle at point \(P(1, -1)\) (where the two lines intersect) with centre \(C\left(\dfrac{4}{3}, \dfrac{1}{3}\right)\) is:</p>
<p>\(x + 4y + 3 = 0\)</p>
<p>\(x - 4y + 3 = 0\)</p>
<p>\(4x + y + 3 = 0\)</p>
<p>\(4x - y - 3 = 0\)</p>
Step-by-Step Solution
Key Concept: The tangent at any point on a circle is perpendicular to the radius at that point. Find the slope of radius CP, then use the perpendicularity condition (m₁·m₂ = -1) to find the tangent's slope.
<p><strong>Step 1:</strong> Verify P(1, -1) lies on both given lines.</p><p>Line 1: 1 - (-1) = 2 ≠ 1 ✗ | Line 2: 2(1) + (-1) = 1 ≠ 3 ✗</p><p>Note: The problem states these are the given lines; we proceed with finding the tangent at P with center C.</p><p><strong>Step 2:</strong> Find the slope of radius CP.</p><p>Slope of CP = $\frac{\frac{1}{3} - (-1)}{\frac{4}{3} - 1} = \frac{\frac{1}{3} + 1}{\frac{4}{3} - 1} = \frac{\frac{4}{3}}{\frac{1}{3}} = 4$</p><p><strong>Step 3:</strong> Find the slope of tangent using perpendicularity.</p><p>If slope of radius = 4, then slope of tangent = $-\frac{1}{4}$ (since $m_1 \cdot m_2 = -1$)</p><p><strong>Step 4:</strong> Write the equation of tangent at P(1, -1).</p><p>Using point-slope form: $y - (-1) = -\frac{1}{4}(x - 1)$</p><p>$y + 1 = -\frac{1}{4}x + \frac{1}{4}$</p><p>$4y + 4 = -x + 1$</p><p>$x + 4y + 3 = 0$</p><p>∴ Answer: A</p>
Correct Answer: A