Circles
Radical axis / common chord
Grade 11
Question:
<p>If the circles \(x^2 + y^2 + 5kx + 2y + k = 0\) and \(2(x^2 + y^2) + 2kx + 3y - 1 = 0\), \((k \in R)\), intersect at the points P and Q, then the line \(4x + 5y - k = 0\) passes through P and Q, for</p>
<p>infinitely many values of \(k\).</p>
<p>no value of \(k\).</p>
<p>exactly two values of \(k\).</p>
<p>exactly one value of \(k\).</p>
Step-by-Step Solution
Key Concept: The radical axis (line through intersection points P and Q) is obtained by subtracting one circle equation from the other. This radical axis must be identical to the given line 4x + 5y - k = 0, allowing us to find k by comparing coefficients.
<p><strong>Step 1: Rewrite both circles in standard form</strong></p><p>Circle 1: $x^2 + y^2 + 5kx + 2y + k = 0$</p><p>Circle 2: $2(x^2 + y^2) + 2kx + 3y - 1 = 0$ → $x^2 + y^2 + kx + \frac{3}{2}y - \frac{1}{2} = 0$</p><p><strong>Step 2: Find the radical axis by subtracting Circle 2 from Circle 1</strong></p><p>$(x^2 + y^2 + 5kx + 2y + k) - (x^2 + y^2 + kx + \frac{3}{2}y - \frac{1}{2}) = 0$</p><p>$5kx - kx + 2y - \frac{3}{2}y + k + \frac{1}{2} = 0$</p><p>$4kx + \frac{1}{2}y + k + \frac{1}{2} = 0$</p><p><strong>Step 3: Multiply by 2 to eliminate fractions</strong></p><p>$8kx + y + 2k + 1 = 0$</p><p><strong>Step 4: Compare with the given line 4x + 5y - k = 0</strong></p><p>For these to represent the same line, the coefficients must be proportional:</p><p>$\frac{8k}{4} = \frac{1}{5} = \frac{2k+1}{-k}$</p><p><strong>Step 5: Solve using the second and third ratios</strong></p><p>$\frac{1}{5} = \frac{2k+1}{-k}$ → $-k = 5(2k+1)$ → $-k = 10k + 5$ → $-11k = 5$ → $k = -\frac{5}{11}$</p><p>∴ Answer: B</p>
Correct Answer: B