Question

find the equation of circle which has center on the line y=2 and which passes through the points (2,0)and (4,0)?

Answer 1

here c is (x,2)
radius will be equal
so,after squaring,
(x-2)^2+(2-0)^2=(x-4)^2+(2-0)^2
4x=12
x=3
so c is (3,2)
hence equation is (x-3)^2+(y-2)^2=r^2
{now r^2 will be (3-2)^2+(2-0)^2=5}
so eq is x^2+y^2-6x-4y+8=0
hope it will help you....