<s>[INST] shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
 [/INST] Let $c$ be a real number such that $3x+4y=c$ has common points $(x,y)$ with the circle and parallel to $3x+4y=12$.
Thus, by C-S we obtain:
$$|c|=|3x+4y|\leq\sqrt{(3^2+4^2)(x^2+y^2)}=\sqrt{25\cdot4}=10,$$ which says that
$3x+4y=10$ is a tangent to the circle, 
which is parallel to the line $3x+4y=12$ and  nearest to this line.
Id est, the needed distance it's:
$$\frac{|12-10|}{\sqrt{3^2+4^2}}=\frac{2}{5}.$$
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