# Circle graphs

• Jan 6th 2011, 03:24 AM
jgv115
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).
• Jan 6th 2011, 03:40 AM
alexmahone
Quote:

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?
• Jan 6th 2011, 03:43 AM
DrSteve
Recall the standard form for the equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

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.