**How to graph circles ? **

Here we are going to see how to graph circles.

There are two types of circles,

(i) Circles with center at origin (0, 0)

(ii) Circles with center at some point (h, k)

**Case 1 :**

If the given equation is in the form

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

then the center of the circle will be at origin.

Here "r" stands for radius

**Case 2 :**

If the given equation is in the form

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

then center of the circle will be at the point (h, k)

**Case 3 :**

If the given equation is in the form

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

then center of the circle will be at the point (-g, -f)

radius = √(g^{2} + f^{2} - c)

Let us look into some example problems to understand the above concepts.

**Example 1 :**

Graph the circle given below

x^{2 }+ y^{2} = 16

**Solution :**

Since the given equation is in the form x^{2} + y^{2} = r^{2}, center of the circle is at the point (0, 0)

Here, r^{2} = 16

r = √16 ==> 4 units

**Example 2 :**

Graph the circle given below

(x - 2)^{2 }+ (y + 3)^{2} = 16

**Solution :**

Since the given equation is in the form (x-h)^{2}+(y-k)^{2} = r^{2}, center of the circle is at the point (h, k)

(x - 2)^{2 }+ (y - (-3))^{2} = 4^{2}

Center (h, k) ==> (2, -3)

radius = 4

**Example 3 :**

Graph the circle given below

x^{2 }+ y^{2} - 2x - 6y + 1 = 0

**Solution :**

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

x^{2 }+ y^{2} - 2x - 6y + 1 = 0 ---(1)

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

By comparing the (1)^{st} and (2)^{nd} equations, we get

2g = -2 ===> g = -1

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

Center (-g, -f) ==> (1, 3)

Radius = √(g^{2} + f^{2} - c)

= √(1^{2} + 3^{2} - 1)

= √(1 + 9 - 1)

= √9 ==> 3 units

After having gone through the stuff given above, we hope that the students would have understood "How to graph circles".

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