Equation of circle after rolling Q: A circle of radius $5$ units touches both the axes and lies in the first quadrant. If the circle makes one complete roll on $+ x$-axis, then what will be its equation in the new position $?$
I got my answer as $x^2 +y^2 -x(10+20\pi)-10y+100\pi^2+100\pi+25=0.$
Is it correct?  

The circle of radius $5$ has equation $(x-5)^2+(y-5)^2=5^2$.
One complete revolution covers a distance of $2\pi r$, or $10\pi$ in this case. 
So the centre also shifts by $10\pi$ to the right.
So the new equation is
$$(x-5-10\pi)^2+(y-5)^2=5^2$$
So:
$$x^2-2(5+10\pi)x+(5+10\pi)^2+y^2-10y+25=25$$
$$x^2-(10+20\pi)x+25+100\pi+100\pi^2+y^2-10y=0$$
Thus:
$$x^2+y^2-x(10+20\pi)-10y+100\pi^2+100\pi+25=0$$
So your proof is correct. 
