We will learn how to find the equation of a circle passes through the origin and centre lies on y-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 = 0 and k = a.
Then the equation (x - h)\(^{2}\) + (y - k)\(^{2}\) = a\(^{2}\) becomes x\(^{2}\) + (y - a)\(^{2}\) = a\(^{2}\)
If a circle passes through the origin and centre lies on y-axis then the y co-ordinate will be equal to the radius of the circle and the abscissa of the centre will be zero. Hence, the equation of the circle will be of the form:
x\(^{2}\) + (y - a)\(^{2}\) = a\(^{2}\)
⇒ x\(^{2}\) + y\(^{2}\) - 2ay = 0
Solved example on
the central form of the equation of a circle passes through the origin and
centre lies on y-axis:
1. Find the equation of a circle passes through the origin and centre lies on y-axis at (0, -6).
Solution:
Centre of the lies on x-axis at (0, -6)
Since, circle passes through the origin and centre lies on y-axis then the y co-ordinate will be equal to the radius of the circle and the abscissa of the centre will be zero.
The required equation of the circle passes through the origin and centre lies on y-axis at (0, -6) is
x\(^{2}\) + (y + 6)\(^{2}\) = (-6)\(^{2}\)
⇒ x\(^{2}\) + y\(^{2}\) + 12y + 36 = 36
⇒ x\(^{2}\) + y\(^{2}\) + 12y = 0
2. Find the equation of a circle passes through the origin and centre lies on y-axis at (0, 20).
Solution:
Centre of the lies on y-axis at (0, 20)
Since, circle passes through the origin and centre lies on y-axis then the y co-ordinate will be equal to the radius of the circle and the abscissa of the centre will be zero.
The required equation of the circle passes through the origin and centre lies on y-axis at (0, 20) is
x\(^{2}\) + (y - 20)\(^{2}\) = 20\(^{2}\)
⇒ x\(^{2}\) + y\(^{2}\) - 40y + 400 = 400
⇒ x\(^{2}\) + y\(^{2}\) - 40y = 0
● The Circle
11 and 12 Grade Math
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