How do you write an equation of a circle with center (3,0) passes through the point (-2,4)?

1 Answer
Dec 26, 2016

#x^2+y^2-6x-32=0#

Explanation:

WE know that the equation of a circle with center #(a,b)# and radius #r# is #(x-a)^2-(y-b)^2=r^2#.

Here although center is given as #(3,0)#, its radius has not been given explicitly. But as circle passes through #(-2,4)# and radius is the distance between center and any point on the circumference, the radius is equal to distance between #(3,0)# and #(-2,4)#, which is

#sqrt((3-(-2))^2-(0-4)^2)=sqrt((3+2)^2-(-4)^2)=sqrt(25+16)=sqrt41#

Hence equation of circle is

#(x-3)^2-(y-0)^2=(sqrt41)^2#

or #x^2-6x+9+y^2=41#

or #x^2+y^2-6x-32=0#
graph{x^2+y^2-6x-32=0 [-16, 16, -8, 8]}