Thread: Circle graphs

I have a basic understanding of circular graphs but I don't know how to do this question

Find the equation of the circle whose centre lies on the line y = 4 and which passes through the points (2, 0) and (6, 0).

Originally Posted by jgv115

I have a basic understanding of circular graphs but I don't know how to do this question

Find the equation of the circle whose centre lies on the line y = 4 and which passes through the points (2, 0) and (6, 0).

The centre of a circle lies on the perpendicular bisector of any 2 points on its circumference. The perpendicular bisector of (2, 0) and (6, 0) is the line x = 4. But we also know that the centre lies on the line y = 4. So, the centre of the circle is (4, 4).

Can you proceed?

Recall the standard form for the equation of a circle:



where (h,k) is the center of the circle and r is the radius of the circle.

Now use the information from Alex's post.