Math Doubts

Equation of a circle when centre of the circle lies on x-axis

Equation

${(x-a)}^2 + y^2 = r^2$
$x^2 + y^2 -2ax + a^2 -r^2 = 0$

Proof

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 $C(a, 0)$.

centre of the circle lies on x axis

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 $QCP$.

  1. Length of the opposite side is $PQ = y$
  2. Length of the adjacent side is $CQ = OQ \,–\, OC = x-a$
  3. 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.

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

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

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

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