Expert: Josh - 1/10/2009

Question
Find the equation of a circle whose diameter runs from (-5,7) to the point of intersection of the lines 2x-y =7 and x+y=2. Express your answer in standard form.

I know that the equation of a circle is (x-h)^2 + (y-k)^2 = r^2.

Subbing in the info, (-5-h)^2 + (7-k)^2=r^2...unsure what to do next?

Answer
Anan,

You are sort of half way there. The problem you are facing is that you need to work out the diameter (2R) or radius (R) from the information given to you.  One way to look at the problem is that the diameter is given by distance AC, where A and C represent the end points of the diameter on the boundary of the circle. If we draw a straight line between A and C, its midpoint B=(h,k) is the center of the circle.

We know that A=(-5,7). To find C=(p,q), we solve the two line equations 2x-y=7 ...[1] and x+y=2 ...[2] simultaneously to arrive at the solution of x=p and y=q. You can use any standard algebra technique: e.g., process of elimination (adding [1] and [2] to work out x and y in turn) or direct substitution of one variable into the other equation (make y the subject in [2] and put this in [1]) to work out x and y.

Once you know the coordinates of A=(r,s) and C=(p,q), you can use Pythagoras theorem to find the Euclidean distance. i.e., the radius R is given by |AC|/2, where |AC|=sqrt((r-p)^2+(s-q)^2). [I should clarify that r represents the x-coordinate of point A here, I have used R to denote the radius]

From coordinate geometry, use the formula for the midpoint: viz., B=(t,u) where t=(r+p)/2, u=(s+q)/2 to find the center of the circle.

Finally, it's a trivial matter putting the value of h,k,R into the equation (x-h)^2+(y-k)^2=R^2.

Can I leave you to have a go at this yourself? Let me know if you have any problems.

Regards
Josh