**Definition of circle:**

Circle is the two dimensional closed geometrical figure in which the distance from the center point to the locus of all points is constant.

**Domain of the circle** --- all the possible x values are said to be domain of the circle.

**Range of the circle**----- all the possible y values are said to be range of the circle.

The standard form of the circle is

`(x- a)^2` +`(y-b)
^2` =r^{2}

Here

r is the radius of the circle

(a, b) is the center of the circle.

**The formula to find the domain of circle: **

The domain will be {x| a-r < x < a+r}

**The formula to find the range of the circle:**

The range will be {y| b-r < y < b+r}

1.Find the domain and range of the circle whose equation is x^{2}+y^{2}-16=0

**Solution:**

Rewrite the equation to standard form.

**The standard form of the circle is**

` (x- a)^2` +`(y-b)
^2` =r^{2}

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

x^{2}+y^{2}=
4^{2}

Here

r is the radius of the circle,4

(a, b) is the center of the circle,(0,0)

**The formula to find the domain of circle: **

The domain will be {x| a-r < x < a+r}

Domain is {x|0-4 <x<0+4}

Domain is {x|-4<x<4}

**The formula to find the range of the circle:**

The range will be {y| b-r < y < b+r}

Range is {y| 0-4 <y<0+4}

Range is {y|-4<y<4}

2.find the domain and range of the circle whose equation is (x-2)^{2}+(y-4)^{2}-49=0

**Solution:**

Rewrite the equation to standard form.

**The standard form of the circle is**

`(x- a)^2` +`(y-b)
^2` =r^{2}

So, x^{2}+y^{2}=7^{2}

x^{2}+y^{2}=
7^{2}

Here

r is the radius of the circle,7

(a, b) is the center of the circle,(2,4)

**The formula to find the domain of circle: **

The domain will be {x| a-r < x < a+r}

Domain is {x|2-7 <x<2+7}

Domain is {x|-5<x<9}

**The formula to find the range of the circle:**

The range will be {y| b-r < y < b+r}

Range is {y| 4-7 <y<4+7}

Range is {y|-3<y<11}