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?
On
The circle travels a distance of $2\pi\cdot 5=10\pi$ units to the right after one full rotation. It's original equation is $(x-5)^2+(y-5)^2=5^2$. Hence, its new position is $$(x-5-10\pi)^2+(y-5)^2=5^2\iff (x-5(1+2\pi))^2+(y-5)^2=25\\\iff x^2-10(1+2\pi)x+25(1+2\pi)^2+y^2-10y+25=25\\\iff x^2+y^2-x(10+20\pi)-10y+100\pi^2+100\pi+25=0$$ So yes, you're 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.