In the xy-plane, an equation of circle A is x^2+(y-1)^2=1. Circle B is obtained by shifting
circle A one unit down. Which of the following is an
equation of circle B?
A) x^2+y^2=1
B) x² + (y-2)² = 1
C) (x - 1)^2 +(y – 1)^2 = 1
D) (x + 1)^2 + (y – 1)^2 = 1

Please with explanation

A

Explanation:

Because the center of circle A is (0,1) and circle B is one unit down, it would be x^2+y^2=1 because the center of circle B is (0,0). (The equation for a circle is (x-h)^2+(y-k)^2=r^2 with h and k as coordinates of the center of the circle and r as the radius.)

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.

okay wait before i answer, is this for your points on a graph?