Centre of the Circle on y-axis

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

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The Circle