Need help: The equations of three circles are x^2 + y^2 - 2x - 4y -4 = 0, x^2 + y^2 - 8x + 4y + 16 = 0 and 4x^2 + 4y^2 - 48x - 16y + 135 = 0. Find the number of points common to at least two of the circles.
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Tiffi1984
ANSWERS: 1
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If you divide the fourth equation by four, all equations have x^2+y^2 the next two terms divided by -2 give the center of the circle: -2x-4y means the center is at x1 y2 By rewriting the equation in the form: (x-1)^2 + (y-2)^2 = r^2 you can find the radius. The first equation is: ((x-1)^2 - (1)^2) + ((y-2)^2 - (2)^2) - 4 = 0 (x-1)^2 + (y-2)^2 = 4 + (1)^2 + (2)^2 = 9 radius is 3. The second equation leads to center (4,-2) radius 2 The distance between the centers is (3,-4) which obviously has length 5. The cases are: If r1+r2 = d, circles touch externally, one point in common (the case with the first two circles) If |r2-r1| = d, circles touch with the smaller touching the inside of the bigger, one point in common If |r2-r1| > d, one circle is entirely within the other, no points in common. If r1+r2 < d, the circles don't come close enough to touch, no points in comon. In all other cases the circles intersect in two common points. To finish the question, repeat between the other two pairs of circles.
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