Find the shortest distance between $9x^2+9y^2-30y+16=0$ and $y^2=x^3$ Find the shortest distance between $9x^2+9y^2-30y+16=0$ and $y^2=x^3$.
I know the shortest distance exists between the curves on the common normal line. Is there any other shorter way to attempt?

As the first curve is a circle you only have to find its center, and than compute the normal equations of your second curve. the one which goes through the center of the circle will do the job. 
$$
\begin{align*}9x^2 + 9y^2 -30y+16 &= 9 \left( x^2+y^2 - \frac{10}{3}y +\frac{16}{9}\right)\\
&=9 \left(x^2+y^2-2\cdot \frac{5}{3} +\left(\frac{5}{3}\right)^2 -\left(\frac{5}{3}\right)^2 +\frac{16}{9}\right)\\
&=9\left(x^2 +\left( y- \frac{5}{3}\right)^2 -1\right)=0
\end{align*}$$
So the center of the circle wil be $$\begin{pmatrix} 0 \\ \frac{5}{3}\end{pmatrix}$$
