Question

unit 4 transformations > lesson 3 translations as functions
ricardo draws a circle on the coordinate plane. the circle has the equation $sqrt{(16-(x + 2)^2)}-5$. the circle is now translated 3 units to the right and 3 units up. where is the translated circle’s center?
(1 point)

Explanation:

Step1: Rewrite equation in standard form

The general form of a circle equation is $(x - a)^2+(y - b)^2=r^2$. First, rewrite $y=\sqrt{16-(x + 2)^2}-5$ as $(x + 2)^2+(y + 5)^2=16$. The center of the original circle is $(-2,-5)$.

Step2: Apply translation rules

For a translation of 3 units to the right and 3 units up, we add 3 to the x - coordinate and 3 to the y - coordinate of the center. The new x - coordinate is $-2+3 = 1$, and the new y - coordinate is $-5 + 3=-2$.

Answer:

$(1,-2)$