Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$

I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original post hinted towards, setting $k = \frac{y+1}{x+2}$ and than writing the given equation as $$(x-3)^2 + (k(x+2) - 4)^2 = 6.$$ I than proceeded to expand and simplify, which gave me $$x^2(k^2 + 1) + x(4k^2 - 8k - 6) + (4k^2 - 16k + 19).$$ However, I am unsure where to go from here. Should I use the discriminant now?

The function and the constraint are
$$
f(x,y) = \frac{y+1}{x+2}\\
\quad g(x,y) = (x-3)^2 + (y-3)^2 - 6 = 0
$$
using Lagrange multipliers
$$
\nabla f = (-\frac{y+1}{(x+2)^2}, \frac{1}{x+2}), \quad \nabla g = (2x,2y)
$$
We get the equations
$$
\nabla f - \lambda \nabla g = 0\\
g(x,y) = 0
$$
in $\lambda, x$, and $y$. The first (vector) equation is the two equations
$$
-\frac{y+1}{(x+2)^2} - \lambda 2x = 0  \tag{1} 
$$
$$
\frac{1}{x+2} - \lambda 2y = 0 \tag{2}
$$
Mathematica gives me the solution set
$$
x = \frac{1}{41}(93-4\sqrt{210}),\quad y = \frac{1}{41}(99+5\sqrt{210}),\quad \lambda = \frac{8400+971\sqrt{210}}{151620} \\
x = \frac{1}{41}(93+4\sqrt{210}),\quad y = \frac{1}{41}(99-5\sqrt{210}),\quad \lambda = \frac{8400-971\sqrt{210}}{151620} \\
$$
