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{x2+y2+6x−12y+20=0y2=4x\begin{cases} { x }^{ 2 }+{ y }^{ 2 }+6x-12y+20=0 \\ { y }^{ 2 }=4x \end{cases} {x2+y2+6x−12y+20=0y2=4x
The shortest distance between the above curves can be written in the form ab−c a\sqrt { b } -c ab−c, then find a+b+c a+b+c a+b+c.
Note: a a a,b bb and c c c are positive integers and b b b is not a perfect square.
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