If largest and smallest value of $\frac{y - 4}{x - 3}$ is $p$ and $q$ where $(x, y)$ satisfy $x^2 + y^2 - 2x - 6y + 9 = 0$ then which of the following is true:
Step-by-Step Solution
Key Concept: Tangent line from external point satisfies both the tangent condition and the point condition simultaneously.
The tangent to circle $(x-1)^2 + (y-3)^2 = 1$ has form $y - 3 = m(x-1) + \sqrt{1+m^2}$. Since point $(3,4)$ lies on this tangent, we get $1 = 2m + \sqrt{1+m^2}$, which simplifies to $(1-2m)^2 = 1+m^2$. Expanding yields $3m^2 - 4m = 0$, giving $m = 0$ or $m = \frac{4}{3}$. The smallest slope is $0$ and the largest is $\frac{4}{3}$.
Correct Answer: 1,3