Circles
Chord of Contact
Grade 11
Question:
<p>The distance between the chords of contact of tangents to the circle \(x^2 + y^2 + 2gx + 2fy + c = 0\) from the origin and the point \((g, f)\) is:</p>
<p>(a) \(\sqrt{g^2 + f^2}\)</p>
<p>(b) \(\frac{g^2 + f^2 - c}{2}\)</p>
<p>(c) \(\frac{g^2 + f^2 - c}{2\sqrt{g^2 + f^2}}\)</p>
<p>(d) \(\frac{g^2 + f^2 + c}{2\sqrt{g^2 + f^2}}\)</p>
Step-by-Step Solution
Key Concept: The chord of contact from an external point to a circle is the locus of points whose polars pass through that point. The distance between two parallel chords of contact can be found using the perpendicular distance formula from a point to a line.
<p><strong>Step 1: Find the chord of contact from the origin (0,0).</strong></p><p>For a circle x² + y² + 2gx + 2fy + c = 0, the chord of contact from point (x₁, y₁) is:<br>xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0</p><p>From origin (0,0): x(0) + y(0) + g(x + 0) + f(y + 0) + c = 0<br>⟹ <strong>gx + fy + c = 0</strong> ... (1)</p><p><strong>Step 2: Find the chord of contact from point (g, f).</strong></p><p>From point (g, f):<br>x(g) + y(f) + g(x + g) + f(y + f) + c = 0<br>gx + fy + gx + g² + fy + f² + c = 0<br>⟹ <strong>2gx + 2fy + g² + f² + c = 0</strong> ... (2)</p><p><strong>Step 3: Verify the chords are parallel.</strong></p><p>Equation (1): gx + fy + c = 0<br>Equation (2): 2gx + 2fy + g² + f² + c = 0 can be rewritten as:<br>2(gx + fy) + g² + f² + c = 0</p><p>Both have the same normal direction (g, f), so they are parallel.</p><p><strong>Step 4: Calculate the perpendicular distance between parallel lines.</strong></p><p>Rewrite equation (2) as: gx + fy + (g² + f² + c)/2 = 0</p><p>Distance between lines gx + fy + c = 0 and gx + fy + (g² + f² + c)/2 = 0 is:</p><p>d = |c - (g² + f² + c)/2| / √(g² + f²)<br>= |(2c - g² - f² - c)/2| / √(g² + f²)<br>= |(c - g² - f²)/2| / √(g² + f²)<br>= |-(g² + f² - c)/2| / √(g² + f²)<br>= <strong>(g² + f² - c) / (2√(g² + f²))</strong></p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C