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?

No need to find the intersection, that is no need for quadratics.  The normal to $y=-\frac34x+3$ through the origin is given by the equation $y=\frac43x$; determine the intersection point of both to find that the distance from the line to the origin is $\frac{3}{\sqrt{1+\left(-\frac34\right)^2}}=\frac{12}{5}$.  Hence the distance from the circle to the line is $\frac{12}{5}-2$.
NB: In general the distance from $y=mx+b$ to the origin is $\frac{|b|}{\sqrt{1+m^2}}$
