The general equation for the family of circles is

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

with `(a,b) ` as center and `r` radius

Let us compare our equation with the general equation

there is only one x term, so a becomes 0, i.e., `(x-a)^2 = (x-0)^2 = x^2`

but there are two terms of y. To find b value we need to find the roots of the quadratic polynomial of `y^2 + 10 y` .

One of the ways to find roots of this polynomial is completing square method.

`(y)^2 + 2*y*5 + (5)^2 -(5)^2` (equating this to `a^2 + 2 ab + b^2` identity)

`= (y+5)^2 -25 `

Now combine `x` and `y` terms and write the equation for the circle

`(x-0)^2 + [y- (-5)]^2 -25 = 0 `

`(x-0)^2 +[y -(-5)]^2 = 5^2 `

`therefore` **the center of the circle is (0,-5)**

**and the radius is 5 **

Posted on

Hello!

To find a center and a radius of a circle given by its equation, it is desirable to express this equation in the form

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

Then the center is (a, b) and the radius is r.

In our equation `x` is already of the form `(x-a)^2` with `a=0.` But `y` has two terms, `y^2+10y.` Let's extract a perfect square from it:

`y^2+10y = (y^2+2*5*y+5^2)-5^2 = (y+5)^2-5^2=(y-(-5))^2-5^2.`

And the entire equation becomes

`(x-0)^2+(y-(-5))^2-5^2=0,` or

`(x-0)^2+(y-(-5))^2=5^2.`

So the center is **(0, -5)** and the radius is **5**.

Posted on

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