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?
 A: Your idea is correct. 
You need to redo your calculation of intersects.
You have the  point $(\frac {6}{5}, \frac {6}{5})$ on the line $y=\frac {4}{3} x$ which does not make sense. 
A: 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 $\frac{|b|}{\sqrt{1+m^2}}$
A: 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 $-\sqrt{6^2+8^2}\le 6\cos t+8\sin t\le?$
A: Let $$x_c^2+y_c^2=4$$ and $$3x_s+4y_s=12$$ then $$d=\sqrt{(x_c-x_s)^2+(y_c-y_s)^2}$$ you can eliminate two variables.
A: 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
A: 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:
$$\frac{|12-10|}{\sqrt{3^2+4^2}}=\frac{2}{5}.$$
