What is the center of the circle with the equation x2 + y2 +4x-8y+11 = 0 ?

c

Step-by-step explanation:

(-2,4)

Step-by-step explanation:

You do this by completing the square for x and y.

x^2 + 4x + (4/2)^2 +y^2 - 8x + (-8/2)^2 = - 11 + 4 + 16

(x + 2)^2 + (y - 4)^2 = 9

The center should be (-2,4)

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

the correct answer is the second option