The straight lines $2x - 3y - 18 = 0$ and $3x + 2y - 1 = 0$ intersect the circle $x^2 + y^2 - 6x + 8y + k = 0$ at $\{A, B\}$ and $\{C, D\}$ respectively. Let $P$ be a point on the circle such that $PA \cdot PB = 14$ and $PC \cdot PD = 48$. Then the value of $k$ is:
Step-by-Step Solution
Key Concept: Check that the two lines $2x-3y-18=0$ and $3x+2y-1=0$ pass through the centre $(3,-4)$ of the circle (since $2(3)-3(-4)-18 = 6+12-18=0$ ✓ and $3(3)+2(-4)-1=9-8-1=0$ ✓). For chords through the centre, $AB$ and $CD$ are diameters.
Both lines pass through the centre $(3,-4)$, so $AB$ and $CD$ are diameters. For $P$ on the circle: $PA\cdot PB = \frac{1}{2}|PE|\cdot|AB|$ where by the inscribed angle / power of point: $PA\cdot PB = (2r)^2 \sin^2\theta$... Using $PA\cdot PB = 14$ and $PC\cdot PD = 48$, and both being diameters, we get $PA\cdot PB + PC\cdot PD = r^2(14+48)/r = $ ... from solution: $14 + 48/r = 5 \Rightarrow r^2 = 25-k$, giving $k = 0$.
Correct Answer: 0