Question

The circle $x^2+y^2-2x-2y+1=0$ is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in es new-position.

Answer 1

Given circle $x^2+y^2-2x-2y+1=0$
Rewriting the equation, we get,
$x^2-2x+1+y^2-2y+1=1(x-1{)}^2+(y-1{)}^2=1\dots \left(1\right)$
The given circle has its centre at (1, 1) and radius = 1 from (1). When circle is rolled on x-axis, it centre moves horizontally through distance = $2\pi$

Figure shows circle with centre (1, 1) at P. After rolling it on X-axis, it takes the position Q.
The coordinates of it's centre become $(1,1+2\pi )$.
Radius of the circle at Q = 1
Hence, equation of new circle is $[x-(1+2\pi {)}^2]+(y-1{)}^2$