Intersection of two orthogonal parabolas

Geometry Level 3

Consider the following two parabolas which intersect each other such that their axes of symmetry form a right angle: { y = x 2 − 3 x = 2 y 2 − 4. \begin{cases} y = x^2 - 3 \\ x = 2y^2 - 4. \end{cases} (You can plot these two curves using the graphing tool in www.desmos.com.)

Interestingly, all four points of their intersection lie on a common circle.

Find the radius of this circle.


The answer is 2.3045.

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Vijay Simha by Vijay Simha , 47, USA

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1 solution

Hosam Hajjir
Oct 17, 2017

Suppose a point ( x , y ) (x, y) is an intersection between the two curves, then

y = x 2 − 3 y = x^2 - 3

and

x = 2 y 2 − 4 x = 2 y^2 - 4

Dividing the second equation by 2 2 , gives

1 2 x = y 2 − 2 \dfrac{1}{2} x = y^2 - 2

Adding this to the first equation,

y + 1 2 x = x 2 + y 2 − 5 y + \dfrac{1}{2} x = x^2 + y^2 - 5

Re-arranging,

x 2 − 1 2 x + y 2 − y = 5 x^2 - \dfrac{1}{2} x + y^2 - y = 5

Completing the square for both x and y , results in,

( x − 1 4 ) 2 + ( y − 1 2 ) 2 = 5 + 1 16 + 1 4 = 5.3125 = r 2 (x - \dfrac{1}{4})^2 + (y - \dfrac{1}{2})^2 = 5 + \dfrac{1}{16} + \dfrac{1}{4} = 5.3125 = r^2

Which is the equation of a circle with center ( 1 4 , 1 2 ) ( \dfrac{1}{4} , \dfrac{1}{2} ) and radius r = 5.3125 = 2.30489 r = \sqrt{5.3125} = 2.30489

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