The equation of a circle when the circle touches the y-axis is written in one of the following two forms in mathematics.

$(1).\,\,\,$ $(x-a)^2+(y-b)^2 \,=\, a^2$

$(2).\,\,\,$ $(x-h)^2+(y-k)^2 \,=\, h^2$

Let us learn the geometric procedure to know how to prove the equation of a circle in mathematical form when the circle is touching the $y$-axis in two dimensional space.

A right triangle (or right angled triangle) should has to construct geometrically inside a circle when the circle touches the $y$-axis at a particular point in a quadrant of the two dimensional Cartesian coordinate system for deriving the equation of circle.

- Draw a circle in such a way that it touches the vertical $y$-axis at a point in the first quadrant of two dimensional cartesian coordinate system.
- Let $C$ be a point, which represents the center (or centre) of the circle. Suppose, the centre $C$ is $a$ and $b$ units away from origin in the horizontal and vertical directions. Hence, the center $C$ in coordinate form is written as $C(a, b)$ mathematically.
- Let $P$ be a point on the circumference of the circle and the point $P$ is $x$ units away from the origin in horizontal direction, and is also $y$ units away from the origin in vertical direction. Therefore, the point $P$ in coordinate form is written as $P (x, y)$ in mathematics.
- Join the point $P$ and centre $C$ by a straight line. The length of the line segment that joins the both points $C$ and $P$ is the radius of the circle, and it is taken as $r$ units. Draw a straight line horizontally from center $C$ and also draw a line vertically from point $P$ towards $x$-axis in such a way that they both got intersected at a point $Q$. Geometrically, a right triangle $(\Delta PCQ)$ is constructed.

In $\Delta PCQ$, $\overline{PQ}$, $\overline{CQ}$ and $\overline{CP}$ are the opposite side, adjacent side and hypotenuse, and the lengths of them are written as $PQ$, $CQ$ and $CP$ respectively in mathematics.

Now, let us find the length of every side of the right angled triangle in algebraic form.

- $CQ$ $\,=\,$ $OQ\,–\,OC$ $\,=\,$ $x \,–\,a$
- $PQ$ $\,=\,$ $OP\,–\,OQ$ $\,=\,$ $y\,-\,b$
- $CP \,=\, r$.

In this right triangle, the length of each side is known to us. So, the relation between three sides can be written in mathematical form by using the Pythagorean Theorem.

${CP}^2$ $\,=\,$ ${CQ}^2+{PQ}^2$

Now, replace the length of each side in the above equation.

$r^2$ $\,=\,$ $(x \,-\, a)^2$ $+$ $(y \,-\, b)^2$

$\implies$ $(x-a)^2$ $+$ $(y-b)^2$ $\,=\,$ $r^2$

Here, $OC \,=\, r \,=\, a$.

$\,\,\,\therefore\,\,\,\,\,\,$ $(x-a)^2$ $+$ $(y-r)^2$ $\,=\,$ $r^2$

This equation can also be written as follows.

$\,\,\,\therefore\,\,\,\,\,\,$ $(x-a)^2$ $+$ $(y-b)^2$ $\,=\,$ $a^2$

The above two equations express the equation of a circle when the circle is touching the vertical $y$-axis at a point.

Use the square of difference formula to expand the square of difference of the terms.

$\implies$ $x^2$ $+$ $a^2$ $-$ $2ax$ $+$ $y^2$ $+$ $b^2$ $–$ $2by$ $\,=\,$ $a^2$

$\implies$ $x^2$ $+$ $y^2$ $-$ $2ax$ $–$ $2by$ $+$ $a^2$ $+$ $b^2$ $\,=\,$ $a^2$

$\implies$ $x^2$ $+$ $y^2$ $-$ $2ax$ $–$ $2by$ $+$ $b^2$ $+$ $a^2$ $\,=\,$ $a^2$

$\implies$ $x^2$ $+$ $y^2$ $-$ $2ax$ $–$ $2by$ $+$ $b^2$ $+$ $a^2$ $-$ $a^2$ $\,=\,$ $0$

$\implies$ $x^2$ $+$ $y^2$ $-$ $2ax$ $–$ $2by$ $+$ $b^2$ $+$ $\cancel{a^2}$ $-$ $\cancel{a^2}$ $\,=\,$ $0$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^2$ $+$ $y^2$ $-$ $2ax$ $–$ $2by$ $+$ $b^2$ $\,=\,$ $0$

This equation represents the circle’s equation when the circle touches the $y$-axis at a point and it can also be written in the following form by replacing the $a$ with $r$.

$\,\,\,\therefore\,\,\,\,\,\,$ $x^2$ $+$ $y^2$ $-$ $2rx$ $–$ $2by$ $+$ $b^2$ $\,=\,$ $0$

The equation of a circle when the circle is touching the $y$-axis is also written in the following two forms if the centre (or center) is considered as $C (h, k)$.

$(x-h)^2+(y-k)^2 \,=\, h^2$

$x^2$ $+$ $y^2$ $-$ $2hx$ $–$ $2ky$ $+$ $k^2$ $\,=\,$ $0$

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