The center of a circle is on the right focus of the ellipse 16x ^ 2 + 25y ^ 2 = 400, and passes through the vertex of the ellipse on the y-axis to find the equation of the circle

The center of a circle is on the right focus of the ellipse 16x ^ 2 + 25y ^ 2 = 400, and passes through the vertex of the ellipse on the y-axis to find the equation of the circle

The center of the circle can be obtained. Then the distance formula between two points can be obtained. The radius can be obtained. Finally, the equation can be written

Given that the circle passes through two focal points of the ellipse x ^ 2 / 4 + y ^ 2 = 1D, and taking the vertex of the ellipse in the positive direction of the Y axis as the center of the circle, the equation of the circle is obtained?
It's x ^ 2 / 4 + y ^ 2 = 1

x^2/4+y^2=1
a^2=4,b^2=1,c^2=4-1=3
The vertex of the ellipse in the positive direction of Y axis is (0,1), which is the center of the circle
Radius R ^ 2 = C ^ 2 + B ^ 2 = 3 + 1 = 4
Circle square x ^ 2 + (Y-1) ^ 2 = 4

Taking the right focus F of the ellipse x24 + Y23 = 1 as the center of the circle and passing through the endpoint of the minor axis of the ellipse, the equation is______ ．

From the elliptic equation x24 + Y23 = 1, a = 2, B = 3 are obtained. According to the properties of the ellipse, C = A2 − B2 = 1, so the coordinates of the right focus f are (1,0), that is, the coordinates of the center of the circle are (1,0), and the coordinates of a are (0,3), so the radius of the circle r = (1 − 0) 2 + (0 − 3) 2 = 2

A focal point is perpendicular to the two ends of the minor axis. The eccentricity of the ellipse is——

The angle between Xia ranta and X is 90 △ 2 = 45 degrees
So it's an isosceles right triangle
So B = C
So e ^ 2 = C ^ 2 / (b ^ 2 + C ^ 2) = 1 / 2
e=√2/2