<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] A bit of geometry.
Introduce new coordinates X,Y:
$Y := y+1;$ $X := x+2;$
1)$Y=CX;$
2)$(X-5)^2 + (Y-4)^2=6;$
Need to find the slope $C$ of the line $Y=CX$ s.t. it is tangent to the circle 2).
3)Distance of  tangent line $Y=CX$ to the centre of circle $(5,4):$
$\dfrac{|-5C+4|}{\sqrt{1+C^2}}=√6;$
$(-5C+4)^2=6(1+C^2);$
A quadratic equation for $C$. Choose the larger solution.
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