The equation of a circle can be given in either standard or general form. In order to find the coordinates of its center from the standard form, the following procedure must be followed:

### When the equation is in standard form

The standard form of the equation of a circle is `(x - h)^2 + (y - k)^2 = r^2` where (h, k) are the coordinates of its center. Thus, the numbers which are in place of 'h' and 'k' in the equation of a circle are the coordinates of its center. For example,

#### Example

Find the coordinates of the center of a circle having the equation `(x - 1)^2 + (y - 3)^2 = 4`

On comparing the above equation with the standard form of the equation of a circle `(x - h)^2 + (y - k)^2 = r^2`, we get

h = 1 and k = 3. Thus the coordinates of the center of the circle having the given equation are (1, 3). Let us take a look at another example.

#### Another Example

Find the coordinates of the center of a circle having the equation `(x + 1)^2 + (y + 3)^2 = 4`.

In the above equation, the numbers in place of 'h' and 'k' are not 1 and 3 (as you might think), but they are -1 and -3. That is, h = -1 and k = -3. Why the minus sign? It's because, if you compare the standard form `(x - h)^2 + (y - k)^2 = r^2` with the above equation, you notice that in the standard form there is a minus sign before 'h' and thus it is (x - h) whereas in the equation given above, we have (x + 1). Now we can write (x + 1) as (x - -1) and thus 'h' is -1. Similarly, (y + 3) can be written as (y - -3) thereby giving us k = -3. Thus, the coordinates of the center of the circle having the above equation are (-1, -3).

In the above equation, the numbers in place of 'h' and 'k' are not 1 and 3 (as you might think), but they are -1 and -3. That is, h = -1 and k = -3. Why the minus sign? It's because, if you compare the standard form `(x - h)^2 + (y - k)^2 = r^2` with the above equation, you notice that in the standard form there is a minus sign before 'h' and thus it is (x - h) whereas in the equation given above, we have (x + 1). Now we can write (x + 1) as (x - -1) and thus 'h' is -1. Similarly, (y + 3) can be written as (y - -3) thereby giving us k = -3. Thus, the coordinates of the center of the circle having the above equation are (-1, -3).

Take a look at a few more solved examples here to get it clear, and practice some more questions yourself here to become good at it. Then thank yourself :). Now let us discuss how to get the coordinates of the center when you have the general form of the equation of a circle.

### When the equation is in General Form

The general form of the equation of a circle is `x^2 + y^2 + Cx + Dy + E = 0`. Cutting down all the details, the coordinates of the center of the circle having the above equation are `(C/2, D/2)`. That is, take the coefficient of 'x', divide it by 2, then take the coefficient of 'y', divide it by 2 and you have the coordinates of the center. Take a look at the following example:

#### Example

Find the coordinates of the center of a circle having the equation `x^2 + y^2 - 6x + 8y + 10 = 0`.

Comparing the above equation with the general form `x^2 + y^2 + Cx + Dy + E = 0`, we have C = -6 and D = 8. Dividing these values by 2, we get `-6/2 = -3` and `8/2 = 4`. Thus, the coordinates of the center of the circle having the above equation are (-3, 4).

Note that when you have the equation of the circle in the general form, you take the values 'C' and 'D' with the signs before them, and when you have the equation of the circle in the standard form, you take the values 'h' and 'k' with the signs opposite to the signs before either of them respectively.

You can take a look at a few more solved examples on finding the center from the general form here and practice some questions here.

Thanks for reading the article. Hope you understood it. Please post any comment you might like to make below.

Thanks for reading the article. Hope you understood it. Please post any comment you might like to make below.

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