Question
A circle has its centre on the x-axis and passes through(4,1)and(1,5).find its equation.
Answers
Steve
You know that
(x-h)^2 + (y-k)^2 = r^2 is the equation of a general circle.
Since the center is on the x-axis, k=0, and you have
(x-h)^2 + y^2 = r^2
Plug in your numbers to get
(4-h)^2 + 1 = r^2
(1-h)^2 + 25 = r^2
That gives you
h = -3/2
r^2 = 125/4
so,
(x + 3/2)^2 + y^2 = 125/4
(x-h)^2 + (y-k)^2 = r^2 is the equation of a general circle.
Since the center is on the x-axis, k=0, and you have
(x-h)^2 + y^2 = r^2
Plug in your numbers to get
(4-h)^2 + 1 = r^2
(1-h)^2 + 25 = r^2
That gives you
h = -3/2
r^2 = 125/4
so,
(x + 3/2)^2 + y^2 = 125/4