How do you find general form of circle centered at (1,0) and containing point (-2,3)?

1 Answer
Dec 20, 2015

Answer:

#(x-1)^2+y^2=18#

Explanation:

The general form of a circle:

#(x-h)^2+(y-k)^2=r^2#

where #(h,k)# is the center of the circle and #r# is the radius.

Since the center of the circle is #(1,0)#, we can fill in #h# and #k#.

#(x-1)^2+(y-0)^2=r^2#

Now, all we need to know is the radius length, which we can find by applying to distance formula with the circle's center and another point on the circle: #(-2,3)#.

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#d=sqrt((-2-1)^2+(3-0)^2)#

#d=sqrt18#

Thus, the radius is #sqrt18#, and the radius is represented in the general form of a circle by #r^2#, so #r^2=18#.

Thus, the equation of the circle is:

#(x-1)^2+y^2=18#

graph{(x-1)^2+y^2=18 [-10, 10, -5, 5]}