Expressing a circle in a mathematical form expression when the circle passes through the origin of Cartesian coordinate system and centre of the circle lies on $x$-axis is defined Equation of the Circle when circle passes through the origin and centre of the circle lies on $x$-axis.

Geometrically, a circle can be passed through the origin of the Cartesian coordinate geometric system and also the centre of the same circle may also lie on the horizontal x-axis. It can be expressed in a mathematical equation form in mathematics by understanding the relation of circle with Cartesian coordinate system and it understands us how exactly a circle is, in Cartesian coordinate system.

Consider a circle, whose radius is assumed as $r$ units. Also assumed, the centre of the circle lies on the $x$-axis and the circle passes through the origin of the Cartesian coordinate system.

Assume, the centre of the circle is denoted by a letter $P$ and the point $P$ is located at $a$ units distance from the origin in horizontal $x$-axis direction but the point is lied on the $x$-axis. So, the vertical distance of this point $P$ is zero units. Therefore, the coordinates of the point $P$ is $P(a,0)$.

Consider a point on the circle, called point $R$ which is located at $R(x,y)$. Assume, it is a variable point, which represents each point on the circle.

Draw a line towards horizontal axis from the point $R$ perpendicularly and draw a line from point $P$ on the horizontal axis until both lines get intersected at a point, called $Q$. This procedure constructed a right angled triangle, known $\Delta RPQ$.

The triangle is a right angled triangle. So, applying Pythagorean Theorem to triangle $\Delta RPQ$ gives us relation between three sides of the triangle.

${PR}^{2}={PQ}^{2}+{RQ}^{2}$

The values of the each side of the right angled triangle can be evaluated from the right angled triangle $\Delta RPQ$.

- Length of the Opposite side is $RQ=y$
- Length of the Adjacent side is $PQ=OQ\u2013OP=x\u2013a$
- Length of the Hypotenuse is $PR=r$

Substitute these three values in the mathematical relation, developed from the Pythagorean Theorem.

${r}^{2}={(x\u2013a)}^{2}+{y}^{2}$

$\Rightarrow {(x\u2013a)}^{2}+{y}^{2}={r}^{2}$

The point $P$ is located at $a$ units distance from the origin on the $x$-axis. So, the length of the $OP$ is equal to $a$ units; means $OP=a$. Actually, the point $P$ is centre of the circle and the circle passes through the origin. So, the distance from origin $O$ to point $P$ is known radius of the circle. It means $OP=r$. Therefore, $OP=r=a$. As per this rule, the algebraic expression can be written as follows.

${(x\u2013a)}^{2}+{y}^{2}={a}^{2}$

$\Rightarrow {x}^{2}+{a}^{2}\u20132ax+{y}^{2}={a}^{2}$

$\Rightarrow {x}^{2}\u20132ax+{y}^{2}=0$

It’s an algebraic equation, which represents a circle when the circle is passing through the origin and also its centre lies on the $x$-axis.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.