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
11 and 12 Grade Math
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