A mathematical equation in algebraic form that represents a circle when the circle touches the both horizontal and vertical axes is called equation of a circle when the circle touches the both axes.

In this special case, a circle is passed through a point on horizontal axis and also passed through another point on vertical axis. The circle can be represented in algebraic form by the geometric analysis in the two dimensional Cartesian coordinate system.

The circle’s equation when the circle touches both $x$-axis and $y$-axis can be derived in algebraic form in two different geometric approaches.

1

Image a circle of radius $r$ units. The circle touches the $x$-axis by passing through a point and it is $a$ units distance from the origin in $x$-axis direction. Similarly, it touches the $y$-axis at a point by passing through a point and the point is $b$ units distance from the origin in $y$-axis direction.

Geometrically, the circle’s touching point on $x$-axis and centre of the circle both are at same distance from origin in $x$-axis direction. So, the $x$-coordinate of the centre of the circle is also $a$. Similarly, the circle’s touching point on $y$-axis and centre of the circle both are at same distance from origin in $y$-axis direction. Therefore, the $y$-coordinate of the circle’s centre is also $b$. Hence, the centre of the circle is represented as $C(a, b)$ in coordinate system.

In this case, the distance from circle’s touching point on $y$-axis to centre of the circle is $a$ but it is known as radius of the circle. Therefore, $r = a$. Similarly, the distance from circle’s touching point on $x$-axis to centre of the circle is $b$ but it is also known as radius of the circle. Therefore, $r = b$. Finally, it is cleared that $r = a = b$.

Consider any point on the circle’s circumference and assume it is located at $x$ units distance from origin in $x$-axis direction and also located at $y$ units distance from origin in $y$-axis direction. Therefore, it is represented as $P (x, y)$ in coordinate system.

Join circle’s centre and point $P$. Draw a line from point $C$ and it should be parallel to $x$-axis and also draw another line from $P$ and it should be parallel to $y$-axis. The two lines intersect perpendicularly at point $Q$. The process constructed a right angled triangle $PCQ$ geometrically.

Calculate the lengths of all three sides of the right angled triangle $PCQ$.

- Length of the opposite side ($\overline{PQ}$) is $PQ = OP \,–\, OQ = y \,-\, b$
- Length of the adjacent side ($\overline{CQ}$) is $CQ = OQ \,–\, OC = x \,-\, a$
- Length of the hypotenuse ($\overline{CP}$) is $CP = r$

Use Pythagorean Theorem and write relation between three sides in mathematical form.

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

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

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

In this case, $r = a = b$. Transform the circle in terms of radius of the circle.

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

Expand square of each difference based binomial.

$\implies x^{\displaystyle 2} + r^{\displaystyle 2} \,- 2rx + y^{\displaystyle 2} + r^{\displaystyle 2} \,- 2ry = r^{\displaystyle 2}$

$\implies x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} + r^{\displaystyle 2} \,- r^{\displaystyle 2} = 0$

$\require{cancel} \implies x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} + \cancel{r^{\displaystyle 2}} \,- \cancel{r^{\displaystyle 2}} = 0$

$\therefore \,\, x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} = 0$

It is the equation of the circle in algebraic form when the circle touches both axes and it can be written in the following simple form.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2r(x+y) + r^{\displaystyle 2} = 0$

Substitute $r$ by $a$ to write the circle’s equation in terms of $a$.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2a(x+y) + a^{\displaystyle 2} = 0$

Substitute $r$ by $b$ to write the circle’s equation in terms of $b$.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2b(x+y) + b^{\displaystyle 2} = 0$

2

The equation of circle when the circle is touched the both axes can also be derived by calculating the distance between circle’s centre and any point on the circle.

The distance from point $C(a, b)$ to $P(x, y)$ is radius of the circle. So, write the relation between them.

$\sqrt{{(x-a)}^{\displaystyle 2} + {(y-b)}^{\displaystyle 2}} = r$

Square both sides

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

In this case $r = a = b$. Transform the equation in terms of either $a$ or $r$ or $b$ and then expand it to obtain the required equation of the circle.

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

$\implies x^{\displaystyle 2} + r^{\displaystyle 2} \,- 2rx + y^{\displaystyle 2} + r^{\displaystyle 2} \,- 2ry = r^{\displaystyle 2}$

$\implies x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} + r^{\displaystyle 2} \,- r^{\displaystyle 2} = 0$

$\implies x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} + \cancel{r^{\displaystyle 2}} \,- \cancel{r^{\displaystyle 2}} = 0$

$\therefore \,\, x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} = 0$

It can be written in terms of $a$ or $b$.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2ax \,- 2ay + a^{\displaystyle 2} = 0$

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2bx \,- 2by + b^{\displaystyle 2} = 0$

If $a$ and $b$ are $x$ and $y$ coordinates of the centre of the circle and $x$ and $y$ are coordinates of any point on the circumference of the circle, the equation of the circle when the circle touches the both axes, can be written in three different forms.

Equation of circle in terms of radius and coordinates of any point on the circle.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2rx \,- 2ry + r^{\displaystyle 2} = 0$

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2r(x+y) + r^{\displaystyle 2} = 0$

Equation of circle in terms of $x$-coordinate of centre of the circle and coordinates of any point on the circle.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2ax \,- 2ay + a^{\displaystyle 2} = 0$

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2a(x+y) + a^{\displaystyle 2} = 0$

Equation of circle in terms of $y$-coordinate of centre of the circle and coordinates of any point on the circle.

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2bx \,- 2by + b^{\displaystyle 2} = 0$

$x^{\displaystyle 2} + y^{\displaystyle 2} \,- 2b(x+y) + b^{\displaystyle 2} = 0$

Save (or) Share

Copyright © 2012 - 2017 Math Doubts, All Rights Reserved