Solve: $x$, $y$ are real, and $(x-3)^2+(y-3)^2=6$, find the maximum value of $\frac{y+1}{x+2}$
This question is from a class I'm taking, and the current topic is discriminants. I found the discriminant of the equation in x and y to find the maximum and minimum values of x and y that yield real solutions for $(x-3)^2+(y-3)^2=6$, but I can't figure out how to maximize $\frac{y+1}{x+2}$ because if I maximize x, I don't know whether y is minimized. Hints would be appreciated!
Hint Set $\frac{y+1}{x+2}=k$. The problem then asks for the maximum value of $k$.
You have $$(x-3)^2+(y-3)^2=6 \\ y=k(x+2)-1$$ Replacing in the first equation you get $$ (x-3)^2+\bigl(k(x+2)-1-3\bigr)^2=6 $$
This is a quadratic equation for $x$, and if you think for a moment, you should realize that the problem is equivalent to: Find the largest $k$ for which this equation has a real solution.