We will learn how to find the equation of a circle passes through the origin and centre lies on x-axis.
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 x-axis 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 x-axis then the abscissa will be equal to the radius of the circle and the y co-ordinate 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 x-axis:
1. Find the equation of a circle passes through the origin and centre lies on y-axis at (0, -2).
Solution:
Centre of the lies on y-axis at (0, -2)
Since, circle passes through the origin and centre lies on x-axis then the abscissa will be equal to the radius of the circle and the y co-ordinate of the centre will be zero.
The required equation of the circle passes through the origin and centre lies on y-axis 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 x-axis at (12, 0).
Solution:
Centre of the lies on x-axis at (12, 0)
Since, circle passes through the origin and centre lies on x-axis then the abscissa will be equal to the radius of the circle and the y co-ordinate of the centre will be zero.
The required equation of the circle passes through the origin and centre lies on x-axis at (12, 0) is
(x - 12)\(^{2}\) + y\(^{2}\) = 12\(^{2}\)
⇒ x\(^{2}\) - 24x + 144 + y\(^{2}\) = 144
⇒ x\(^{2}\) + y\(^{2}\) - 24x = 0
● The Circle
11 and 12 Grade Math
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