Circles
Tangent to a circle
Grade 11
Question:
<p>Equation of the tangent to the circle, at the point \((1, -1)\), whose centre is the point of intersection of the straight lines \(x - y = 1\) and \(2x + y = 3\) is</p>
<p>\(x + 4y + 3 = 0\)</p>
<p>\(3x - y - 4 = 0\)</p>
<p>\(x - 3y - 4 = 0\)</p>
<p>\(4x + y - 3 = 0\)</p>
Step-by-Step Solution
Key Concept: Find the center by solving the intersection of two lines, then use the property that the tangent at any point on a circle is perpendicular to the radius at that point.
<p><strong>Step 1: Find the center of the circle</strong></p><p>Solve the system: x - y = 1 and 2x + y = 3</p><p>From first equation: x = y + 1</p><p>Substituting: 2(y + 1) + y = 3 → 3y + 2 = 3 → y = 1/3</p><p>So x = 1/3 + 1 = 4/3</p><p>Center C = (4/3, 1/3)</p><p><strong>Step 2: Find slope of radius</strong></p><p>Radius connects C(4/3, 1/3) to point P(1, -1)</p><p>Slope of CP = (-1 - 1/3)/(1 - 4/3) = (-4/3)/(-1/3) = 4</p><p><strong>Step 3: Find slope of tangent</strong></p><p>Since tangent ⊥ radius: m_tangent × 4 = -1</p><p>m_tangent = -1/4</p><p><strong>Step 4: Write equation of tangent</strong></p><p>Using point-slope form at (1, -1):</p><p>y - (-1) = -1/4(x - 1)</p><p>y + 1 = -1/4(x - 1)</p><p>4y + 4 = -x + 1</p><p>x + 4y + 3 = 0</p><p>∴ Answer: C</p>
Correct Answer: C