# Standard Form of a Circle

## Circle

A circle is the path of a point which moves at a constant distance from a fixed point in a plane.  The fixed point is known as center of the circle and the constant distance is known as  the radius of the circle.
Radius of a circle will be always a positive constant.

Look at the figure below.  This is a circle with center O and radius r.

## Standard form of a circle

If the center of the circle is at the point (h, k) and has radius of the circle is r, then the equation of the circle is given by (x - h)2+ (y - k)2= r2
This representation of the circle is called the standard form.

Proof

Let c(h, k) be the center of the circle and r be the radius of the circle. Let L(x, y) be a point on the circle. The CL = r. Draw perpendiculars from CP and LQ to the x axis and and draw perpendicular CM.
We can see that CM = PQ = OQ - OP = x - h
ML = LQ - MQ = LQ - CL = y - k
Δ CML is a right angled triangle.  So by Pythagoras theorem,
CM2 + ML2 = CL
That is (x - h)2 + (y - k)2 = r2
Hence proved.

Lets consider a few examples.

Example 1: -
Write the equation of the circle in the standard form, whose center is (2, 5) and radius is 3.

Solution: -
Given center of the circle,(h, k) = (2, 5) and radius of the circle, r = 3.
We know that the standard form of a circle is (x - h)2 + (y - k)2 = r2
Substituting for h, k and r, we get
(x - 2)2 + (y - 5)2 = 32
Simplifying we get,
(x - 2)2 + (y - 5)2 = 9
So the required equation is (x - 2)2 + (y - 5)2 = 9.

Example 2: -
Write the equation of the circle in the standard form whose center is (3, -2) and radius is 4.
Solution: -
Given center of the circle, (h, k) = (3, -2) and radius of the circle, r = 4.
We know that the standard form of a circle is (x - h)2 + (y - k)2= r2
Substituting for h, k and r, we get
(x - 3)2 + (y - (-2))2 = 42
Simplifying we get,
(x - 3)2 + (y + 2)2 = 16
So the required equation is (x - 3)2 + (y + 2)2 = 16.

Example 3: -
Identify the center and radius of the circle (x + 10)2 + (y - 4)2 = 25.
Solution: -
Given equation is (x + 10)2 + (y - 4)2 = 25.
We know that the standard form of a circle is (x - h)2 + (y - k)2 = r2
Comparing the given equation with the standard form, we get
h = -10, k = 4 and r2 = 25. Taking square root on both sides of r2 = 25, we get r = 5.
So the center of the circle is (-10, 4) and radius is 5.

Example 4: -
Identify the center and radius of the circle (x + 7)2 + (y + 12)2 = 36.
Solution: -
Given equation is (x + 7)2 + (y + 12)2 = 36
We know that the standard form of a circle is (x - h)2 + (y - k)2 = r2
Comparing the given equation with the standard form, we get
h = -7, k = -12 and r2 = 36. Taking square root on both sides of r2 = 36, we get r = 6
So the center of the circle is (-7, -12) and radius is 6.

### Special cases:

1.  When the center is at origin and radius is r, the equation of the circle in standard form is given by x2 + y2 =  r2
2. When the radius of a circle is 0, the equation of the circle in standard form becomes (x - h)2 + (y - k)2 = 0. This can happen only when the circle reduces to a point.  So this circle is called a point circle.

Try yourself: -
1. Write the equation of the circle with center (2, 0) and radius 9.
2. Identify the center and radius of the circle (x + 7)2 + (y + 16)2 = 64.