Compute the area of the circle that passes through all the intersection points of $4x^2 + 11y^2 = 29$ and $x^2 - 6y^2 = 6.$
+128475 Original graph of the problem : https://www.desmos.com/calculator/xyzfrdpqbv
Because of symmetry......we only need one intersection point
4x^2 + 11y^2 = 29 (1)
x^2 - 6y^2 = 6 → -4x^2 + 24y^2 = -24 (2)
Add (1) and (2) and we have
35y^2 = 5
y^2 = 5/35 = 1/7
y = sqrt (1/7)
And
x^2 - 6 (1/sqrt(7))^2 = 6
x^2 - 6/7 = 6
x^2 = 6 + 6/7
x^2 = 48/7
x = sqrt (48/7)
(0,0) will be the center of the circle
The radius = sqrt (48/7 + 1/7) = sqrt( 49/7) = sqrt (7)
The equation of the circle we need = x^2 + y^2 = 7
Area = pi (sqrt (7))^2 = 7 pi units^2
Here's the completed graph : https://www.desmos.com/calculator/of0i8sikec
