$x^2 + {(y-b)}^2 = r^2$
$x^2 + y^2 -2by + b^2 -r^2 = 0$
Consider a circle having radius of $r$ units in two dimensional Cartesian coordinate system. The centre of circle lies on $y$-axis without touching the $x$-axis. The centre of the circle is $b$ units distance from the origin. Therefore, the centre of circle is $C(0, b)$. Take $P$ as a point on the circumference of the circle and its coordinates are $x$ and $y$. Therefore, the point is $P(x, y)$.
Draw a line from point $C$ and it is parallel to horizontal axis and also draw another straight line from point $P$ and it should be parallel to vertical axis. The two lines intersect at point $Q$. Thus, a right angled triangle $PCQ$ is constructed geometrically.
According to right angled triangle $PCQ$.
The lengths of all three sides are known geometrically. Express the relation between them in mathematical form by using Pythagorean theorem.
${CP}^2 = {CQ}^2 + {PQ}^2$
$\implies r^2 = x^2 + {(y-b)}^2$
$\therefore \,\,\,\,\, x^2 + {(y-b)}^2 = r^2$
It is an equation of the circle in general compact form when the centre of the circle lies on the $y$-axis without touching the $x$-axis.
Use expansion of square of difference of two terms to expand this general equation of circle.
$\implies x^2 + y^2 + b^2 -2by = r^2$
$\implies x^2 + y^2 -2by + b^2 = r^2$
$\therefore \,\,\,\,\, x^2 + y^2 -2by + b^2 -r^2 = 0$
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