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?

Hint
$3x=12-4y=4(3-y)$
WLOG any point on the line $P(4m,3-3m)$
$\dfrac43=\dfrac{3-3m-0}{4m-0}$
$m=?$
Can you find the distance of $P$ from the origin
Subtract the length of the radius from the distance
