Question

What is the standard form of the equation of a circle with center at (-3, 1) and through the point (2, 13)?

Answer 1

`(x+3)^2+(y-1)^2=13^2`
(see below for discussion of alternate "standard form")

Explanation:

The "standard form of an equation for a circle" is
`color(white)("XXX")(x-a)^2+(y-b)^2=r^2`
for a circle with center `(a,b)` and radius `r`

Since we are given the center, we only need to compute the radius (using the Pythagorean Theorem)
`color(white)("XXX")r=sqrt((-3-2)^2+(1-13)^2) = sqrt(5^2+12^2) = 13`
So the equation of the circle is
`color(white)("XXX")(x-(-3))^2+(y-1)^2=13^2`

Sometimes what is being asked for is the "standard form of the polynomial" and this is somewhat different.
The "standard form of the polynomial" is expressed as a sum of terms arranged with decreasing degrees set equal to zero.
If this is what your teacher is looking for you will need to expand and rearrange the terms:
`color(white)("XXX")x^2+6x+9+y^2-2y+1=169`

`color(white)("XXX")x^2+y^2+6x-2y-159=0`