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.

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