Question

if a circle passe though point (-4,2) and is tangent to the x-axis at (2,0), determine the coordinates of its center.

Answer 1

Answer:

The centre is at (2, 10).

Explanation:
(x - a)^2 + (y - b)^2 = r^2 is a circle with radius r and centre(a, b).
It passes through the points (-4, 2) and (2, 0), therefore:-
(-4 - a)^2 + (2 - b)^2 = r^2 and
(2 - a)^2 + (-b)^2 = r^2
Now the x coordinate (a) of the centre is 2 (because of the tangent at (2, 0)
So, the equations become:
6^2 + (2 - b)^2 = r^2
b^2 = r^2
--> 36 + (2 -b )^2 -b^2 = 0
--> 36 + 4 + b^2 - 4b - b^2 = 0
--> 4b = 40
--> b = 10.
So the centre is (2, 10).