<p>The length of the common chord of two circles of radii 3 and 4 unit which intersect orthogonally is \(\frac{k}{5}\), then k equals ............</p>
Step-by-Step Solution
Key Concept: Two circles intersect orthogonally when d² = r₁² + r₂². Use the common chord length formula for orthogonal circles.
<p><strong>Solution approach:</strong> When two circles intersect orthogonally, the tangents at the point of intersection are perpendicular. For circles with radii $r_1 = 3$ and $r_2 = 4$ intersecting orthogonally, the distance between centers $d$ satisfies $d^2 = r_1^2 + r_2^2 = 9 + 16 = 25$, so $d = 5$. Using the formula for common chord length in orthogonal circles: $\text{chord length} = \frac{2r_1r_2}{d} = \frac{2 \cdot 3 \cdot 4}{5} = \frac{24}{5}$. Therefore $k = 24$.</p>
Correct Answer: 24