# Determining all ( a , b ) on the unit circle such that 2 x + 3 y + 1 &#

Determining all $\left(a,b\right)$ on the unit circle such that $2x+3y+1\le a\left(x+2\right)+b\left(y+3\right)$ for all $\left(x,y\right)$ in the unit disk
In the middle of another problem, I came up with the following inequality which needed to be solve for $\left(a,b\right)$ :
$2x+3y+1\le a\left(x+2\right)+b\left(y+3\right)$
for all $\left(x,y\right)\in {\mathbb{R}}^{2}$ with ${x}^{2}+{y}^{2}\le 1.$.
Here the solution for $\left(a,b\right)$ must be a subset of the unit circle, and I believe that it is a singleton.
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Yair Boyle
Rewrite $2x+3y+1\le a\left(x+2\right)+b\left(y+3\right)$ as $\left(2-a\right)x+\left(3-b\right)y\le 2a+3b-1$.
Solve for the intersection of the line $\left(2-a\right)x+\left(3-b\right)y=2a+3b-1$ with ${x}^{2}+{y}^{2}=1$ to get that the $x$-coordinates of the intersection are
$\frac{-2{a}^{2}-3ab+5a+6b-2±\sqrt{-\left(b-3{\right)}^{2}\left(3{a}^{2}+12ab+8{b}^{2}-12\right)}}{{a}^{2}-4a+{b}^{2}-6b+13}.$
If this has two real solutions, then the inequality cannot be satisfied for all $\left(x,y\right)$ in the unit circle, so we need to find the intersection of $-\left(b-3{\right)}^{2}\left(3{a}^{2}+12ab+8{b}^{2}-12\right)\le 0$ and ${a}^{2}+{b}^{2}\le 1$. Simultaneously solving these equations, we find that the real solutions for $\left(a,b\right)$ are $\left(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}\right)$ and $\left(-\frac{2}{\sqrt{13}},-\frac{3}{\sqrt{13}}\right)$. It's easy to see that the first solution is valid and the second is invalid - plug in $\left(x,y\right)=\left(0,0\right)$, for instance.