We will learn how to find the equation of a circle passes through the origin and centre lies on xaxis.
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 circle passes through the origin and centre lies on xaxis i.e., h = a and k = 0.
Then the equation (x  h)\(^{2}\) + (y  k)\(^{2}\) = a\(^{2}\) becomes (x  a)\(^{2}\) + y\(^{2}\) = a\(^{2}\)
If a circle passes through the origin and centre lies on xaxis then the abscissa will be equal to the radius of the circle and the y coordinate of the centre will be zero. Hence, the equation of the circle will be of the form:
(x  a)\(^{2}\) + y\(^{2}\) = a\(^{2}\)
⇒ x\(^{2}\) + y\(^{2}\)  2ax = 0
Solved example on the central form of the equation of a circle passes through the origin and centre lies on xaxis:
1. Find the equation of a circle passes through the origin and centre lies on yaxis at (0, 2).
Solution:
Centre of the lies on yaxis at (0, 2)
Since, circle passes through the origin and centre lies on xaxis then the abscissa will be equal to the radius of the circle and the y coordinate of the centre will be zero.
The required equation of the circle passes through the origin and centre lies on yaxis at (0, 2) is
(x + 7)\(^{2}\) + y\(^{2}\) = (7)\(^{2}\)
⇒ x\(^{2}\) + 14x + 49 + y\(^{2}\) = 49
⇒ x\(^{2}\) + y\(^{2}\) + 14x = 0
2. Find the equation of a circle passes through the origin and centre lies on xaxis at (12, 0).
Solution:
Centre of the lies on xaxis at (12, 0)
Since, circle passes through the origin and centre lies on xaxis then the abscissa will be equal to the radius of the circle and the y coordinate of the centre will be zero.
The required equation of the circle passes through the origin and centre lies on xaxis at (12, 0) is
(x  12)\(^{2}\) + y\(^{2}\) = 12\(^{2}\)
⇒ x\(^{2}\)  24x + 144 + y\(^{2}\) = 144
⇒ x\(^{2}\) + y\(^{2}\)  24x = 0
11 and 12 Grade Math
From Circle Passes through the Origin and Centre Lies on xaxis 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.
