We will learn how to find the equation when the centre of a circle 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 centre of a circle is on the x-axis i.e., k = 0.
Then the equation (x - h)\(^{2}\) + (y - k)\(^{2}\) = a\(^{2}\) becomes (x - h)\(^{2}\) + y\(^{2}\) = a\(^{2}\) ⇒ x\(^{2}\) + y\(^{2}\) - 2hx + h\(^{2}\) = a\(^{2}\) ⇒ x\(^{2}\) + y\(^{2}\) - 2hx + h\(^{2}\) – a\(^{2}\) = 0
If the centre of a circle be on the x-axis, then the y co-ordinate of the centre will be zero. Hence, the general form of the equation of the circle will be of the form x\(^{2}\) + y\(^{2}\) + 2gx + 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
x-axis:
1. Find the equation of a circle whose centre of a circle is on the x-axis at -5 and radius is 9 units.
Solution:
Radius of the circle = 9 units.
Since, centre of a circle be on the x-axis, then the y co-ordinate of the centre will be zero.
The required equation of the circle whose centre of a circle is on the x-axis at -5 and radius is 9 units is
(x + 5)\(^{2}\) + y\(^{2}\) = 9\(^{2}\)
⇒ x\(^{2}\) + 10x + 25 + y\(^{2}\) = 81
⇒ x\(^{2}\) + y\(^{2}\) + 10x + 25 - 81 = 0
⇒ x\(^{2}\) + y\(^{2}\) + 10x - 56 = 0
2. Find the equation of a circle whose centre of a circle is on the x-axis at 2 and radius is 3 units.
Solution:
Radius of the circle = 3 units.
Since, centre of a circle be on the x-axis, then the y co-ordinate of the centre will be zero.
The required equation of the circle whose centre of a circle is on the x-axis at 2 and radius is 3 units is
(x - 2)\(^{2}\) + y\(^{2}\) = 3\(^{2}\)
⇒ x\(^{2}\) - 4x + 4 + y\(^{2}\) = 9
⇒ x\(^{2}\) + y\(^{2}\) - 4x + 4 - 9 = 0
⇒ x\(^{2}\) + y\(^{2}\) - 4x - 5 = 0
● The Circle
11 and 12 Grade Math
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