**Find the center of a circle :**

Here we are going to see how to find centre and radius of circle.

Generally equation of circle will be in the following forms

(i) Center at origin (0, 0) and radius "r"

x^{2} + y^{2} = r^{2}

(ii) Center at (h, k) and radius "r"

(x - h)^{2} + (y - k)^{2} = r^{2}

(iii) Equation of a circle will be in the form

x2 + y2 + 2gx + 2fy + c = 0

If the given equation of circle is in the above form, then center of the circle will be at (-g, -f) and radius is √g^{2} + f^{2} - c

Let us look in to some example problems to understand how to find center and radius of circle.

**Example 1 :**

Find the center and radius of the following circle

x^{2} + y^{2} = 1

**Solution :**

Since the given equation is in the form x^{2} + y^{2} = r^{2 }the center of the circle will be at origin.

Instead of r^{2 }we have 1, Hence the radius of the circle is 1 unit.

**Example 2 :**

Find the center and radius of the following circle

x^{2} + y^{2} - 4x - 6y + 9 = 0

**Solution :**

Since the given equation is in the form x^{2} + y^{2} + 2gx + 2fy + c = 0^{ }the center of the circle will be at (-g, -f)

x^{2} + y^{2} - 4x - 6y + 9 = 0

x^{2} + y^{2} + 2gx + 2fy + c = 0

2g = -4 ==> g = -2

2f = -6 ==> f = -3

(-g, -f) ==> (2, 3)

Radius = √2^{2} + 3^{2} - 9

= √4 + 9 - 9

= √4 = 2 units

**Example 3 :**

Find the center of the circle described on the line joining the points (1, 2) and (2, 4) as its diameter.

**Solution :**

Midpoint of the diameter = Center of the circle

By finding the midpoint of the line segment joining the points, we get the center of the circle.

Midpoint = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

= (1 + 2)/2 , (2 + 4)/2

= (3/2, 6/2)

= (3/2, 3)

After having gone through the stuff given above, we hope that the students would have understood "Find the centre of a circle"

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