Question

A circle is tangent to the y-axis at the point (0,2) and passes through the point (8,0). Find the radius of the circle.

I tried using distance formula but it doesnt work? Help please thanks.

Answer 1

since the line y = 2 is tangent to the circle, its centre must lie on the y-axis

let the centre be (0,b)

the equation of the circle is
x^2+ (y-b)^2 = r^2

but (0,2) lies on it ----> (2-b)^2 = r^2

but (8,0) lies on it ----> 64 + (0-b)^2 = r^2
64 +b^2= r^2

then 64 + b^2 = (2-b)^2
64 + b^2 = 4-2b+b^2

60 = -2b
b = -30

the centre is (0, -30) ---> radius = 2-(-30) = 32

OR

distance from (0,b) to (0,2) must equal the distance from (0,2) to (8,0)

2-b = √(64 + b^2)
square both sides
4-2b+b^2 = 64+b^2
-2b = 60
b = -30
the radius is 2 - (-30) = 32

Answer 2

That is super wrong.