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
From Circle Passes through the Origin to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Nov 29, 23 01:15 AM
Nov 29, 23 12:54 AM
Nov 29, 23 12:50 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.