Question

The equation of auxiliary circle of the ellipse is

• A. x^2 + y^2 - 2x – 4y - 5 = 0
• B. x^2 + y^2 - 2x – 4y - 20 = 0
• C. x^2 + y^2 + 2x + 4y - 20 = 0
• D. x^2 + y^2 + 2x + 4y - 5 = 0

Answer 1

Answer: (B)

x^2 + y^2 - 2x – 4y - 20 = 0

Answer 2

The center of the ellipse is the midpoint of the foci $F_1(-1, 0)$ and $F_2(3, 4)$, which is $C(1, 2)$. The foot of the perpendicular from a focus to a tangent lies on the auxiliary circle. Thus, the point $P(4, 6)$ lies on the auxiliary circle. The radius of the auxiliary circle is the distance $CP$. $a = CP = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. The equation of the auxiliary circle with center $(1, 2)$ and radius $5$ is $(x-1)^2 + (y-2)^2 = 5^2$, which expands to $x^2 + y^2 - 2x - 4y - 20 = 0$.