How do find the standard equation of the circle with center (3,5) and passes through (-1,-2)?

1 Answer
May 4, 2017

Answer:

The standard Cartesian form for the equation of a circle is:

#(x-h)^2+(y-k)^2=r^2" [1]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center, and r is the radius.

Explanation:

Substitute the center #(3,5)# into equation [1]:

#(x-3)^2+(y-5)^2=r^2" [2]"#

Substitute the point #(-1,-2)# into equation [2] and solve for r:

#(-1-3)^2+(-2-5)^2=r^2#

#(-4)^2+(-7)^2=r^2#

#16+49=r^2#

#r^2=65#

#r = sqrt65#

Substitute #sqrt65# for r into equation [2]:

#(x-3)^2+(y-5)^2=(sqrt65)^2" [3]"#

Equation [3] is the answer.

Note: Please understand that I write the right side as #(sqrt65)^2#, instead of 65, because anyone asked for the radius will not make the mistake of reporting 65.