# how do i write the equation of the circle in standard form. I have to find the center,radius and intercept then graph the circle. x^2+y^2-4x+18y=-69

To find the coordinates of the center complete the square for both x and y.

To do this halve the coefficients in front of the terms in x and y which gives -2 and 9 respectively. Considering the expressions

`(x-2)^2 = x^2 - 4x + 4` and `(y+9)^2 = y^2 + 18y + 81`

we can see that the first 2 terms on the lefthand sides of these two equations give us the terms in `x^2, x, y^2` and `y` in the original equation

`x^2 + y^2 - 4x +18y = -69`

Rearranging the expressions,

`(x-2)^2 - 4 = x^2 - 4x` and `(y+9)^2 - 81 = y^2 + 18y`

Substituting these two expressions into the original equation gives

`(x-2)^2 - 4 + (y+9)^2 - 81 = -69`

Collect all the integer terms onto the righthand side, giving

`(x-2)^2 + (y+9)^2 = -69 + 4 + 81`

So we have

`(x-2)^2 + (y+9)^2 = 16` in **standard form** **answer**

At any point `(x,y)` on the circle we can form a right-angled triangle from that point to the centre of the circle `(x_c, y_c )` .

The hypotenuse is the radius of the circle and the horizontal and vertical sides have length `(x-x_c)` and `(y-y_c)` respectively

So using Pythagoras' Theorem `(x-x_c)^2 + (y-y_c)^2 = r^2`

This is exactly the standard form we have obtained above. We can see that setting

`x_c = 2` , `y_c = -9` and `r^2 = 16`

So the **centre** is at `(x_c,y_c) = (2, -9)` and the **radius** is `r = sqrt(16) = 4`

**answer**

To draw the graph on a computer, convert to polar coordinates

`x(theta) = rcos(theta) = 4cos(theta)`

`y(theta) = rsin(theta) = 4sin(theta)`

` `where `theta` is the angle the hypotenuse (radius) makes from the horizontal line

`y = y_c = -9`