Math Doubts

Equation of a circle when the circle touches the both axes

Equation

$(1).\,\,$ $(x-a)^2+(y-a)^2$ $\,=\,$ $a^2$
$(2).\,\,$ $(x-b)^2+(y-b)^2$ $\,=\,$ $b^2$
$(3).\,\,$ $(x-r)^2+(y-r)^2$ $\,=\,$ $r^2$

Introduction

A circle that touches both horizontal and vertical axes of two dimensional Cartesian coordinate system can be expressed in mathematical form by an equation and it is called the equation of a circle when the circle is touching the both axes.

circle equation in standard form

Let the coordinates of center (or centre) of a circle are denoted by $a$ and $b$, and the geometric coordinates of a point on the circumference of a circle are denoted by $x$ and $y$, and the radius of circle is denoted by $r$. Then, the equation of a circle, which touches the both horizontal $x$ axis and vertical $y$ axis of a two dimensional space is written as follows.

Simple form

circle touches both axes equation

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

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

$(3).\,\,$ $(x-r)^2+(y-r)^2$ $\,=\,$ $r^2$

Expansion

$(1).\,\,$ $x^2$ $+$ $y^2$ $-$ $2a(x+y)$ $+$ $a^2$ $\,=\,$ $0$

$(2).\,\,$ $x^2$ $+$ $y^2$ $-$ $2b(x+y)$ $+$ $b^2$ $\,=\,$ $0$

$(3).\,\,$ $x^2$ $+$ $y^2$ $-$ $2r(x+y)$ $+$ $r^2$ $\,=\,$ $0$

Other form

The equation of a circle, which touches the both horizontal axis and vertical axis is also written in the following form by taking $C(h, k)$ as the center or centre of a circle in coordinate form.

Simple form

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

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

$(3).\,\,$ $(x-r)^2+(y-r)^2$ $\,=\,$ $r^2$

Expansion

$(1).\,\,$ $x^2$ $+$ $y^2$ $-$ $2h(x+y)$ $+$ $h^2$ $\,=\,$ $0$

$(2).\,\,$ $x^2$ $+$ $y^2$ $-$ $2k(x+y)$ $+$ $k^2$ $\,=\,$ $0$

$(3).\,\,$ $x^2$ $+$ $y^2$ $-$ $2r(x+y)$ $+$ $r^2$ $\,=\,$ $0$

Proof

Learn how to derive the equation of a circle that touches the both horizontal and vertical axes.

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