Circles
Tangent to a circle
Grade 11
Question:
<p>The equation of circle is \(x^2 + y^2 + 4x - 4y + 4 = 0\). The equation of the tangent to this circle which makes equal intercepts on the positive coordinate axes is:</p>
<p>\(x + y = 2\sqrt{2}\)</p>
<p>\(x + y = -2\sqrt{2}\)</p>
<p>\(x + y = \pm 2\sqrt{2}\)</p>
<p>\(x - y = \pm 2\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: A line making equal intercepts on positive coordinate axes has form x + y = a (where a > 0). Substitute this into the tangent condition: distance from center to line equals radius.
<p><strong>Step 1: Find circle's center and radius</strong></p><p>Rewrite x² + y² + 4x - 4y + 4 = 0 in standard form:</p><p>(x + 2)² + (y - 2)² = 4 + 4 - 4 = 4</p><p>Center C = (-2, 2), Radius r = 2</p><p><strong>Step 2: Form equation of tangent with equal intercepts</strong></p><p>A line making equal intercepts a on positive x and y axes is:</p><p>x/a + y/a = 1, or <strong>x + y = a</strong> (where a > 0)</p><p><strong>Step 3: Apply tangency condition</strong></p><p>Distance from center (-2, 2) to line x + y - a = 0 must equal radius 2:</p><p>|(-2) + 2 - a|/√(1² + 1²) = 2</p><p>|-a|/√2 = 2</p><p>|a| = 2√2</p><p><strong>Step 4: Select positive value</strong></p><p>Since we need positive intercepts: a = 2√2</p><p>∴ The equation of tangent is <strong>x + y = 2√2</strong> or <strong>x + y - 2√2 = 0</strong></p>
Correct Answer: C