How do you write the equation of the circle with center (2,4) and containing the point (-2,1)?

1 Answer
Apr 27, 2016

Answer:

#25=(x-2)^2+(y-4)^2#

Explanation:

The equation of a circle with center #(h,k)# and radius #r# is:
#r^2=(x-h)^2+(y-k)^2#

We are told that the center of this circle is #(2,4)#, so
#r^2=(x-2)^2+(y-4)^2#

However, we don't know the radius of this circle. Luckily, we are also told that this circle contains the point #(-2,1)#, so we will use that information to find #r#:
#r^2=(x-2)^2+(y-4)^2#
#r^2=(-2-2)^2+(1-4)^2#
#r^2=16+9#
#r^2=25#
#r=5#

The radius of the circle is #5#, which means the equation of the circle is:
#25=(x-2)^2+(y-4)^2#