Question

Find the center and radius of the circle whose equation is x^2+ y^2-20 4x- 2y .

Answer 1

is it x^2 -4x + y^2 +2y-20 = 0
(x-2)^2 (y+1)^2 -20 = 0 + 2 + 1
x^2-4x+4 y^2-2y+1-20 =3
x^2-4x + y^2 -2y +5-20
x^2-4x +y^2-2y-15=3
x(x-4) y(y-2)-15=3 is this the right direction so far?

Answer 2

you need to complete the square for terms in x and y

the standard form is

(x -h)^2 + (y = k)^2 = r^2

the centre of the circle is (h, k) and the radius is r

so you have

x^2 - 4x + (4/2)^2 + y^2 + 2y + (2/1)^2 = 20 + (4/2)^2 + (2/1)^2

they terms in x and y can be factorised into perfect squares and the right hand side will give r^2 when simplified.

Answer 3

so it should be (x+2)^2 + (y+1)^2 = 23

Answer 4

not quite

the right hand side should be 20 + 4 + 1

the term in x has a negative middle term so the sign inside the perfect square should be negative...

Answer 5

im srry my brain is crashing been studying for a while. I see the mistake with the negative and kinda with the 20+4+1 but my completing the square is rusty isnt it taking 1/2 of b and adding it on both sides then bringing it down to rewrite it so when its squared it gives u the statement above?

Answer 6

ok... so you have

(x - 2)^2 + (y + 1)^2= 25

use the information above to find the centre and radius

Answer 7

well the radius is 5 and the center is 2,-1 but Im still not seeing how u got 25 lol

Answer 8

I keep getting 23

Answer 9

cause half of b for the y part should be 1

Answer 10

ok... complete the square for

x^2 - 4x b = -4 the constant term in the perfect square is (-4/2)^2

so you are really doing (-2)^2 = 4

add it to both sides of the equation to keep it in balance...

same for

y^2 + 2y b = 2 (2/2)^2 = 1

so add 1 to both sides to keep the equation in balance

so you have

x^2 -4x + 4 + y^2 +2y + 1 = 20 + 4 + 1

(x -2)^2 + (y + 1)^2 = 25

hope that makes sense