Correct this for me arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.

Remark

I'm going to complete the square very quickly for all of these. You can fill in the missing steps.

Top One

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

(x - 1)^2 + (y + 1)^2 = 3

Radius = sqrt(3)

Second One

Divide by 4 to start with

x^2 + y^2 + 4x + 6y - 10 = 0

(x^2 + 4x + 4) + (y^2 + 6y + 9) = 10 + 4 + 9

(x + 2)^2 + (y + 3)^2 = 23

Radius = sqrt(23)

Three

(x^2 - 4x + 4) + (y^2 + 4y + 4) = 10 + 8

(x - 2)^2 + (y + 2)^2 = 18

Radius = sqrt(18) You would be wise not to reduce this even though the reducation is 3sqrt(2). The comparison will be easier.

Four

(x^2 - 8x + 16) + (y^2 - 6y + 9) = 20 + 16 + 9

(x - 4)^2 + (y - 4)^2 = 44

Radius = sqrt(44)

Again don't reduce this

Five

(x^2 - 12x + 36) + (y^2 - 2y + 1) = 9 + 36 + 1

(x -6)^2 + (y - 1)^2 = 46

radius = sqrt(46)

Six

Divide through by 5

x^2 + y^2 - 4x + 6y + 8 = 0

(x^2 - 4x + 4) + (y^2 + 6y + 9) = - 8 + 4 + 9 = 5

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

Radius = sqrt(5)

Seven

Divide through by 2

x^2 + y^2 - 14x - 32y - 4 = 0

(x^2 - 14x + 49) + (y^2 - 32y + 256) = 4 + 49 + 256

(x - 7)^2 + (y - 16)^2 = 309

The radius is sqrt(309)

Comment

The calculation order produces

One . . . . . sqrt(3)

Two . . . . . . sqrt(23)

Three . . . . .sqrt(18)

Four . . . . . .sqrt(44)

Five . . . . . .sqrt(46)

Six . . . . . . .sqrt(5)

Seven . . . . sqrt(309)

I leave the ordering part to you.

answer: y=0 if that is what u r looking for

yes

step-by-step explanation:

yes