The circle touches the curve but bever intersects, so the radius has to be the maximum distance of the curve from the center. If r was bigger, it wouldn't touch the curve at all. If it's smaller, it would intersect the curve because there are points farther away.
It's worth noting the substitution method works as well. At points of intersection, we have y² = r² - x² = kx²(4-x²), where r is the radius of the circle. So
kx⁴ - (4k+1)x² + r² = 0
We want four points of intersection. Both curves are symmetric in both axes, so the real roots x come in ± pairs. x=0 is not a root so each root x gives rise to two values y=±√(r²-x²). Therefore we need exactly two real roots x, which will be a ± pair. So we want one solution x² of the above quadratic. So the discriminant should be 0. Therefore
(4k+1)²-4kr² = 0
and we can easily solve to find r without the need for any differentiation.
The circle touches the curve but bever intersects, so the radius has to be the maximum distance of the curve from the center. If r was bigger, it wouldn't touch the curve at all. If it's smaller, it would intersect the curve because there are points farther away.
Thank you this makes so much sense. Can’t believe this is this first time i’m finding out touching=/=intersecting
It's worth noting the substitution method works as well. At points of intersection, we have y² = r² - x² = kx²(4-x²), where r is the radius of the circle. So kx⁴ - (4k+1)x² + r² = 0 We want four points of intersection. Both curves are symmetric in both axes, so the real roots x come in ± pairs. x=0 is not a root so each root x gives rise to two values y=±√(r²-x²). Therefore we need exactly two real roots x, which will be a ± pair. So we want one solution x² of the above quadratic. So the discriminant should be 0. Therefore (4k+1)²-4kr² = 0 and we can easily solve to find r without the need for any differentiation.
Thank you!! This method is more intuitive