<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] Another way:
Using coordinate geometry,
Any point on the circle $P(2\cos t,2\sin t)$
Distance of $P$ from the line $$\dfrac{|3(2\cos t)+4(2\sin t)-12|}5$$
Now $-\sqrt{6^2+8^2}\le 6\cos t+8\sin t\le?$
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