Question

write the equation of the circle in standard form. x^2+y^2-10x-4y+13=0.
First, i brought 13 over to the left side of the equation, and then grouped terms.
(x^2-10x+___)+(y^2-4y+___)=-13
Then, i completed the square by dividing the numbers before the x's in half, and then squaring them to get...
(x^2-10x+25)+(y^2-4y+4)=-13+25+4
Simplify...
(x-5)^2+(y-2)^2=16
at this point, your equation is in standard form.
r^2=16,so r=4
the center is at (h,k) which is (5,2)
and your intercepts are going to be when you set x=0 and y=0 and solve for each.
For y, -5+(y-2)=4 y-2=9
y=11 (0,11)x=(7,0). Something is wrong! help.

Answer 1

your center and radius are correct, so nice work on getting that

Answer 2

x-intercepts:

(x-5)^2+(y-2)^2=16

(x-5)^2+(0-2)^2=16

(x-5)^2+(-2)^2=16

(x-5)^2+4=16

(x-5)^2=16-4

(x-5)^2=12

x-5 = sqrt(12) or x-5 = -sqrt(12)

x-5 = 2*sqrt(3) or x-5 = -2*sqrt(3)

x = 5 + 2*sqrt(3) or x = 5 - 2*sqrt(3)

Those are your exact x-intercepts. You would use a calculator to approximate them

Answer 3

y-intercepts

(x-5)^2+(y-2)^2=16

(0-5)^2+(y-2)^2=16

25 + (y-2)^2 = 16

(y-2)^2 = 16 - 25

(y-2)^2 = -9

y - 2 = sqrt(-9) or y - 2 = -sqrt(-9)

since the square root of a negative number doesn't produce a real number, this means that there are no real solutions to (0-5)^2+(y-2)^2=16

so there are no y-intercepts

Answer 4

here is the graph of the circle

Answer 5

thank you so much! you're awesome.

Answer 6

you're welcome