Question

Which of the following is a result of shifting a circle with equation (x+3)^2+(y-2)^2=36 up 3 units

Answer 1

Answer:

The general equation of the circle is given by:

$(x-h)^2+(y-k)^2=r^2$ where r is the radius of the circle and (h, k) is the center of the circle.

To shift any function upwards b units , then the center of the circle becomes;

(h, k+b)

As per the given statement:

equation of the circle is given by:

$(x+3)^2+(y-2)^2=36$

Center of this circle is: (-3, 2) and radius(r) = 6

to shift this equation of circle 3 units up then;

(h, k+3) = (-3, 2+3) = (-3, 5)

then;

the equation of circle 3 units upwards with radius = 6 units and center = (-3, 5) become:

$(x+3)^2+(y-5)^2=36$

Therefore, the result of shifting a circle with equation

$(x+3)^2+(y-2)^2=36$   up 3 units is:

 $(x+3)^2+(y-5)^2=36$

Answer 2

Answer: A result of shifting a circle with equation (x+3)^2+(y-2)^2=36 up 3 units is: (x+3)^2+(y-5)^2=36

To shift a graph "a" units up, we must replace in the equation "y" by "y-a" when y is not isolated; or if y is isolated, you add "a" to the right side of the equation.
In this case we want to shift the graph of the circle 3 units up, then a=3, and we must replace in the equation of the circle "y" by "y-3":
(x+3)^2+[(y-3)-2]^2=36→
(x+3)^2+(y-3-2)^2=36→
(x+3)^2+(y-5)^2=36

Answer: A result of shifting a circle with equation (x+3)^2+(y-2)^2=36 up 3 units is: (x+3)^2+(y-5)^2=36