# Circle Passes through the Origin

We will learn how to form the equation of a circle passes through the origin.

The equation of a circle with centre at (h, k) and radius equal to a, is (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$.

When the centre of the circle coincides with the origin i.e., a$$^{2}$$ = h$$^{2}$$ + k$$^{2}$$

Let O be the origin and C(h, k) be the centre of the circle. Draw CM perpendicular to OX.

In triangle OCM, OC$$^{2}$$ = OM$$^{2}$$ + CM$$^{2}$$

i.e., a$$^{2}$$ = h$$^{2}$$ + k$$^{2}$$.

Therefore, the equation of the circle (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ becomes

(x - h)$$^{2}$$ + (y - k)$$^{2}$$ = h$$^{2}$$ + k$$^{2}$$

⇒ x$$^{2}$$ + y$$^{2}$$ - 2hx – 2ky = 0

The equation of a circle passing through the origin is

x$$^{2}$$ + y$$^{2}$$ + 2gx + 2fy = 0 ……………. (1)

or, (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = h$$^{2}$$ + k$$^{2}$$ …………………………. (2)

We clearly see that the equations (1) and (2) are satisfied by (0, 0).

Solved examples on the central form of the equation of a circle passes through the origin:

1. Find the equation of a circle whose centre is (2, 3) and passes through the origin.

Solution:

The equation of a circle with centre at (h, k) and passes through the origin is

(x - h)$$^{2}$$ + (y - k)$$^{2}$$ = h$$^{2}$$ + k$$^{2}$$

Therefore, the required equation of the circle is (x - 2)$$^{2}$$ + (y - 3)$$^{2}$$ = 2$$^{2}$$ + 3$$^{2}$$

⇒ x$$^{2}$$ - 4x + 4 + y$$^{2}$$ – 6y + 9 = 4 + 9

⇒ x$$^{2}$$ + y$$^{2}$$ - 4x – 6y = 0.

2. Find the equation of a circle whose centre is (-5, 4) and passes through the origin.

Solution:

The equation of a circle with centre at (h, k) and passes through the origin is

(x - h)$$^{2}$$ + (y - k)$$^{2}$$ = h$$^{2}$$ + k$$^{2}$$

Therefore, the required equation of the circle is (x + 5)$$^{2}$$ + (y - 4)$$^{2}$$ = (-5)$$^{2}$$ + 4$$^{2}$$

⇒ x$$^{2}$$ + 10x + 25 + y$$^{2}$$ – 8y + 16 = 25 + 16

⇒ x$$^{2}$$+ y$$^{2}$$ + 10x – 8y = 0.

The Circle