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

1 Answer
Feb 13, 2016

Answer:

#(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#