\[\begin{cases} { x }^{ 2 }+{ y }^{ 2 }+6x-12y+20=0 \\ { y }^{ 2 }=4x \end{cases} \]

The shortest distance between the above curves can be written in the form \( a\sqrt { b } -c \), then find \( a+b+c \).

**Note:** \( a \),\( b\) and \( c \) are positive integers and \( b \) is not a perfect square.

×

Problem Loading...

Note Loading...

Set Loading...