Circles
Intersection of Circles
Grade 11
Question:
<p>If the circles $x^2 + y^2 + 5Kx + 2y + K = 0$ and $2(x^2 + y^2) + 2Kx + 3y - 1 = 0$, $(K \in \mathbb{R})$, intersect at the points P and Q, then the line $4x + 5y - K = 0$ passes through P and Q, for</p>
<p>(a) no values of K</p>
<p>(b) exactly one value of K</p>
<p>(c) exactly two values of K</p>
<p>(d) infinitely many values of K</p>
Step-by-Step Solution
Key Concept: The common chord of two intersecting circles is obtained by subtracting their equations. For a given line to pass through the intersection points, its equation must be proportional to the equation of the common chord.
<p><strong>Solution:</strong></p><p>Given circles:</p><p>$x^2 + y^2 + 5Kx + 2y + K = 0$ ... (i)</p><p>$2(x^2 + y^2) + 2Kx + 3y - 1 = 0$ ... (ii)</p><p>The common chord (line PQ) is obtained by subtracting one equation from the other.</p><p>From (i): $x^2 + y^2 + 5Kx + 2y + K = 0$</p><p>From (ii): $x^2 + y^2 + Kx + \frac{3}{2}y - \frac{1}{2} = 0$ (dividing by 2)</p><p>Subtracting: $4Kx + \frac{1}{2}y + K + \frac{1}{2} = 0$</p><p>Multiplying by 2: $8Kx + y + 2K + 1 = 0$</p><p>For the line $4x + 5y - K = 0$ to be the common chord, we need:</p><p>$8Kx + y + 2K + 1 \equiv 4x + 5y - K$</p><p>Comparing coefficients: $8K = 4$ and $1 = 5$ (contradiction)</p><p>Since we get $1 = 5$, which is impossible, the line $4x + 5y - K = 0$ cannot pass through P and Q for any value of K.</p><p>∴ Answer is (a) no values of K</p>
Correct Answer: A