Question

Determine the equations of the following circle with its center at the origin and (-2;3) a point of a circle

Answer 1

Answer:

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Answer 2

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.

Circle Passes through the Origin

Circle Passes through the Origin

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).

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