Solving simultaneously, by eliminating y we get

5x² + 2x – 135 = (x – 5)(21x² + 26x - 225)

(x – 5)(5x + 27) = (x – 5)(21x² + 26x - 225)

(x – 5)(21x² + 26x - 225) - (x – 5)(5x + 27) = 0

(x – 5) (21x² + 21x – 252) = 0

(x – 5) (x² + x – 12) = 0

(x – 5)(x + 4)(x – 3) = 0

Hence the x-coordinates of the intersection points are x = 5, x = - 4 and x = 3.

Substituting into the quadratic equation (to keep the calculations as simple as possible) we get

x = 5 and 21y = 125 + 10 – 135 = 0 so the point is (5, 0)

x = -4 and 21y = 80 – 8 – 135 = -63 so the point is (-4, -3)

x = 3 and 21y = 45 + 6 – 135 = - 84 so the point is (3, -4)

Now, to find the centre of the circle you could determine the perpendicular bisectors of two pairs of these points and calculate where they intersect. But you may have noticed that all of these points are a distance of 5 units from the origin! (It is always helpful to spot 3-4-5 patterns 😊)

Therefore all these points must lie on the circle centre (0, 0) with radius 5.

There is no other possible circle – there is a geometrical theorem that any three distinct non-collinear points lie on the circumference of exactly one possible circle.

Challenge 288 : Circle Search

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