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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 = a2.

When the centre of the circle coincides with the origin i.e., a2 = h2 + k2

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

In triangle OCM, OC2 = OM2 + CM2

i.e., a2 = h2 + k2.


Therefore, the equation of the circle (x - h)2 + (y - k)2 = a2 becomes

(x - h)2 + (y - k)2 = h2 + k2

⇒ x2 + y2 - 2hx – 2ky = 0

The equation of a circle passing through the origin is

x2 + y2 + 2gx + 2fy = 0 ……………. (1)

or, (x - h)2 + (y - k)2 = h2 + k2 …………………………. (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 = h2 + k2

Therefore, the required equation of the circle is (x - 2)2 + (y - 3)2 = 22 + 32

⇒ x2 - 4x + 4 + y2 – 6y + 9 = 4 + 9

⇒ x2 + y2 - 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 = h2 + k2

Therefore, the required equation of the circle is (x + 5)2 + (y - 4)2 = (-5)2 + 42

⇒ x2 + 10x + 25 + y2 – 8y + 16 = 25 + 16

⇒ x2+ y2 + 10x – 8y = 0.

 The Circle





11 and 12 Grade Math 

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