We will learn how to find the equation when the centre of a circle 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 centre of a circle is on the y-axis i.e., h = 0.
Then the equation (x - h)\(^{2}\) + (y - k)\(^{2}\) = a\(^{2}\) becomes x\(^{2}\) + (y - k)\(^{2}\) = a\(^{2}\) ⇒ x\(^{2}\) + y\(^{2}\) - 2ky + k\(^{2}\) = a\(^{2}\) ⇒ x\(^{2}\) + y\(^{2}\) - 2ky + k\(^{2}\) - a\(^{2}\) = 0
If the centre of a circle be on the y-axis, then the x co-ordinate of the centre will be zero. Hence, the general form of the equation of the circle will be of the form x2 + y2 + 2fy + c = 0, where g and c are the constants.
Solved examples on
the central form of the equation of a circle whose centre is on the y-axis:
1. Find the equation of a circle whose centre of a circle is on the y-axis at -3 and radius is 6 units.
Solution:
Radius of the circle = 6 units.
Since, centre of a circle be on the y-axis, then the x co-ordinate of the centre will be zero.
The required equation of the circle whose centre of a circle is on the y-axis at -3 and radius is 6 units is
x\(^{2}\) + (y + 3)\(^{2}\) = 6\(^{2}\)
⇒ x\(^{2}\) + y\(^{2}\) + 6y + 9 = 36
⇒ x\(^{2}\) + y\(^{2}\) + 6y + 9 - 36 = 0
⇒ x\(^{2}\) + y\(^{2}\) + 6y - 27 = 0
2. Find the equation of a circle whose centre of a circle is on the y-axis at 4 and radius is 4 units.
Solution:
Radius of the circle = 4 units.
Since, centre of a circle be on the y-axis, then the x co-ordinate of the centre will be zero.
The required equation of the circle whose centre of a circle is on the y-axis at 4 and radius is 4 units is
x\(^{2}\) + (y - 4)\(^{2}\) = 4\(^{2}\)
⇒ x\(^{2}\) + y\(^{2}\) - 8y + 16 = 16
⇒ x\(^{2}\) + y\(^{2}\) – 8y + 16 - 16 = 0
⇒ x\(^{2}\) + y\(^{2}\) - 8y = 0
● The Circle
11 and 12 Grade Math
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