<s>[INST] 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?
 [/INST] Going from what you have  (and avoiding calculus):
$(k^2 + 1)x^2 + (4k^2 -4k-3)x + (k^2 - 16k + 19)= 0$
If we plug this into the quadratic formula:
$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$
When $b^2 - 4ac < 0$ the formula breaks.
Setting $b^2 - 4ac = 0$ will give us the extreme values of $k.$
$(4k^2 - 8k - 6)^2 - 4(k^2+1)(4k^2 - 16k + 19) \ge 0\\
(16k^4 - 64k^3 + (-48+64)k^2 + 96k + 36) - (16k^4-64k^3 + 4(19+4)k^2 -64k + 76)\ge 0\\
-76k^2 + 160k - 40 \ge 0\\
19k^2 - 40k + 10 \le 0\\
$
And now use the quadratic formula again to find the range of $k.$
$\frac {20 - \sqrt {210}}{19}\le k \le \frac {20 + \sqrt {210}}{19}
$
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