${(x-a)}^2 + y^2 = r^2$

$x^2 + y^2 -2ax + a^2 -r^2 = 0$

Imagine a circle in two dimensional Cartesian coordinate system. The radius of the circle is $r$ and its centre lies on the $x$-axis but it does not touch the $y$-axis. The centre of the circle is $a$ units away from origin. Therefore, the $x$ and $y$ coordinates of the centre are $a$ and $0$ respectively. So, the centre point of the circle is $C(a, 0)$.

Consider any point on the circumference of the circle and the coordinates of the point $P$ are $x$ and $y$. So, the point is $P(x, y)$. Join the points $C$ and $P$ by a line and draw a line from $P$ to horizontal axis but it is perpendicular to the $x$-axis and touches it at point $Q$. Thus, a right angled triangle $QCP$ is constructed geometrically.

According to right angled triangle $QCP$.

- Length of the opposite side is $PQ = y$
- Length of the adjacent side is $CQ = OQ \,–\, OC = x-a$
- Length of the hypotenuse $CP = r$

Use Pythagorean Theorem to express the relation between them.

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

$\implies r^2 = {(x-a)}^2 + y^2$

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

It is a circle’s equation in compact form if centre of the circle lies on the $x$-axis without touching the $y$-axis.

It can be expressed in its expansion form by applying square of difference of two terms formula.

$\implies x^2 + a^2 -2ax + y^2 = r^2$

$\implies x^2 + y^2 -2ax + a^2 = r^2$

$\therefore \,\,\,\,\,\, x^2 + y^2 -2ax + a^2 -r^2 = 0$

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved