A circle is a simple shape of Euclidean geometry consisting of those points in a plane which is equidistant from a given point called the center. The common distance of the points of a circle from its center is called its radius.

The standard equation of a circle is given by,

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

Where, r is the radius of the circle.

If the circle has radius one then it is a unit circle.

The equation of a circle with radius r units and center (h, k) is given by the following equation.

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

This is the standard equation of a circle with (h, k) as it's center and with radius "r" in a XY plane.

If the center of the circle is in the origin, then the previous equation becomes,

h = 0, k = 0

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

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

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

If the radius of the circle is 1, then the equation becomes

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

This is the equation of a unit circle with origin as center.

**Cases:**

1) If the center be at the origin and the radius r, then h=0, k=0 and, so the equation of the circle is x^{2}+y^{2} =r^{2}

2) If the origin lies on the circle, then h^{2}+k^{2}=r^{2} and, so the equation of the circle in this case is x^{2}+y^{2}- 2hx-2ky=0

3) If the center lies on the x-axis, then the coordinate of the center is zero k=0, and so the equation of the circle is (x-h)^{2} + y^{2}=r^{2}

4) If the center lies on the x-axis and the origin lies on the circle, then k=0 and h=r, and so the equation of the circle is x^{2}+y^{2}-2rx=0

**To find the radius and the center of a circle we use the following formulas,**

If the equation of a circle is in the form of

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

Here, the center of the circle is (-g, -f)

and the radius of the circle can be found by the formula,

** r = `sqrt(g^2 + f^2 -c)` **

**Ex 1: Find the center and radius of the circle in the equation, x ^{2} + y^{2} - 8x - 10y + 4 = 0**

**Sol:**

The given equation is in the form of circle equation x^{2} + y^{2} + 2gx + 2fy + c = 0

The center of the circle is (-g, -f)

To find (-g, -f)

2g = -8

g = -4

Also, 2f = -10

f = -5

So the center of the circle is (-4, -5)

To find the radius of the circle, we use the formula ** r = ** **`sqrt(g^2 + f^2 -c)`**

We know, g = -4, f = -5, and c = 4

So the radius of the circle, r = `sqrt( (-4)^2 + (-5) ^2 -4)`

r = `sqrt(16+25 -4) `

r = `sqrt(37)`

So the center and radius of the given circle is (-4, -5) and `sqrt(37)` units

**Ex 2: Find the center and radius of the circle x ^{2} + y^{2} + 4x - 9y + 6 = 0**

**Sol:**

The given equation is in the form of circle equation x^{2} + y^{2} + 2gx + 2fy + c = 0

The center of the circle is (-g, -f)

To find (-g, -f)

2g = -4

g = -2

Also, 2f = -9

f = -9/2

So the center of the circle is (-2, -9/2)

To find the radius of the circle, we use the formula ** r = ** **`sqrt(g^2 + f^2 -c)`**

We know, g = -2, f = -9/2, and c = 6

So the radius of the circle, r = **`sqrt((-2)^2 + (-9/2)^2 -6)`**

r = **`sqrt(4+81/4-6)`**

r = `sqrt(18.25)`

r = `sqrt(18.25)` units

So the center and radius of the given circle is (-2, -9/2) and `sqrt(18.25)` units