Question

(30 POINTS) What are the y intercepts of the circle whose equation is (x+2)^2+(y-4)^2=13?

Answer 1

Hello from MrBillDoesMath!

Answer:  Choice C

Discussion:

The y-intercepts are the x values where y = 0.   Substituting y = 0 in the equation gives

(x+2)^2 +  (0-4)^2 = 13 =>

(x+2)^2 + 16 = 13 =>

Subtract 16 from both sides

(x+2)^2 = 13 - 16 = -3     (*)

The square of a real number is greater than or equal to 0. (Not true for complex numbers). The left side of (*) is therefore >= 0 but as also equals -3. This is not possible so there is no solution to the equation. Choice C.

Thank you,

MrB