Question

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

Answer 1

With the center and the point (-6,2) you can deduce the equation of the circle

(x-xo)^2 + (y-yo)^2 = r^2

(x+3)^2 + (y-4)^ = r^2

r^2 is otained from the center and the point (-6,2)

r^2 = (-6 -(-3))^2 + (2-4)^2 = (-6+3)^2 + (-2)^2 = (-3)^2 + 4 = 9 + 4 = 13.

Then the equation of the circle is

 (x+3)^2 + (y-4)^ = 13

Now we subsitute the point (10,4) into that equation and see whether it belongs to it:

(10+3)^2 + (4-4)^2 = 13^2

13^2 ≠ 13, so the point does not belong to the equation.