If a circle has center (0,0) and a point on the circle (-2,-4) write the equation of the circle.?

The general equation of a circle of radius `r` centred on `(a,b)` is:

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

Thus a circle centered on the origin will have an equation of the form:

`x^2+ y^2 =r^2`

Knowing that `(-2,4)` lies on the circle, we have:

`(-2)^2+ (4)^2 =r^2`

`:. r^2 = 4 + 16 = 20`

Thus the equation is

`x^2+ y^2 =20`

Every point on a circle has the same distance from the center. This distance is the radius `r` of the circle.

So, if `(-2,-4)` is a point on the circle, it means that the radius of the circle is the distance between `(0,0)`, the center, and `(-2,-4)`.

To compute the distance between two points `(x_1y_1)` and `(x_2,y_2)`, the formula is

`d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}`

In this case, `(x_1y_1) = (0,0)` and `(x_2,y_2)=(-2,-4)`. So, their distance is

`d = \sqrt{(0-(-2))^2+(0-(-4))^2} = sqrt(4+16)=sqrt(20)`

Now we know the center `(0,0)` and the radius `sqrt(20)` of the circle. When you have this information, you can write the equation as

`(x-x_0)^2+(y-y_0)^2 = r^2`

where `(x_0,y_0)` is the center and `r` is the radius. So, in this case, the equation is

`(x-0)^2+(y-0)^2 = (sqrt(20))^2`

which can be rewritten as

`x^2+y^2 = 20`