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    Find the equation of the circle which passes through the points (1,3) and (5,7), and has its centre on the line y - x = 4.

    Thanks. :confused:

    Erm, this method might be unnecessarily long, but essentially, note that the line joining (1,3) and (5,7) is parallel to y - x = 4. Find the midpoint M of (1,3) and (5,7). Find the equation of the line perpendicular to y - x = 4 and passing through M, and find where it cuts y - x = 4 - this will be the center of the circle. Then you can go on to find the radius and then the equation of the circle.

    Alternatively, let the center of the circle be at (a,b) and the radius be R. As the center lies on the line y = x + 4, then b = a+4. So the equation of the circle is:

     (x-a)^2 + (y-b)^2 = R^2 \iff (x-a)^2 + (y-a-4)^2 = R^2 .

    Substitute the values for x and y at the two points (1,3) and (5,7) and you have simultaneous equations in a and R which you can solve.

    Or (1,3) to (0,4) is "one left and one up"

    mid point of (1,3) - (5,7) is (3,5) so the centre is "one left and one up" from there i.e. (2,6) ?

    Edit: typo

    The quickest way when you have a circle passing through 2 points (A,B) and (C,D) is to realise that the centre is equidistant from both points.

    So if the centre is (X,Y)

    (X-A)^2 + (Y-B)^2 = (X-C)^2 + (Y-D)^2

    This simplifies a lot, and you get easy simultaneous equations as Y-X = 4
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Updated: July 29, 2009


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