Question

Please helppppp!!!!!!!

Answer 1

Answer:

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

Step-by-step explanation:

The general equation for a circle is the following:

$(x-a)^2+(y-b)^2=c$

Where (a, b) is the center of the circle and ($\sqrt{c}$) is the radius of the circle. In order to shift the function to the right, one has to add to the (a) value; to move it to the left one has to subtract from it. Similarly, to move the function up, one has to add to the (b) value, to move it down, one has to subtract from it. Apply this to the given function:

$(x+1)^2+(y-2)^2=9$

The center of this circle is (-1, 2). The circle is shifted to the right by (3) units, and down by (4). Thus the center is also shifted by these values:

(-1, 2)

Right by (3): (-1 + 3) = 2

Down by (4):  (2 - 4) = -2

Substitute back into the function:

$(x-(2))^2+(y-(-2))^2=9\\\\(x-2)^2+(y+2)^2=9$