Question

The equation of a circle is given below.


x^{2}+(y+4)^{2} = 64x


2


+(y+4)


2


=64x, squared, plus, left parenthesis, y, plus, 4, right parenthesis, squared, equals, 64


What is its center?


((left parenthesis


,,comma


))right parenthesis


What is its radius?


If necessary, round your answer to two decimal places.


units

Answer 1

Answer:

(h,k) = (32,-4)

r = 32

Step-by-step explanation:

The general equation for a circle is given by:

$(x-h)^2+(y-k)^2=r^2$   (1)

where (h,k) is the center of the circle and r is the radius.

You have the following equation:

$x^2+(y+4)^2=64x$   (2)

You first need to complete squares in order to obtain an equation of the form (1). Thus, you have that the second term must be in a perfect square trinomial:

2b = 64

b = 32

Then, you have to sum 32^2 and also subtract the same number in the expression (2):

$x^2-64x+(y+4)^2=0\\\\(x^2-64x+32^2)+(y+4)^2-32^2=0\\\\(x-32)^2+(y+4)^2=32^2$

you compare the last result with expression (1) and obtain that the raiuds of the circle is r = 32

Furthermore, the center of the circle is (h,k) = (32,-4)