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).

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- Jan 6th 2011, 03:24 AMjgv115Circle 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 AMalexmahone
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 AMDrSteve
Recall the standard form for the equation of a circle:

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

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.