Question

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths. Tiles x2 + y2 − 2x + 2y − 1 = 0 x2 + y2 − 4x + 4y − 10 = 0 x2 + y2 − 8x − 6y − 20 = 0 4x2 + 4y2 + 16x + 24y − 40 = 0 5x2 + 5y2 − 20x + 30y + 40 = 0 2x2 + 2y2 − 28x − 32y − 8 = 0 x2 + y2 + 12x − 2y − 9 = 0

Answer 1

We will proceed to convert the equations into standard format to determine the solution.

we know that  

The Standard Form Equation of a Circle is equal to

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

where

(h,k) is the center of the circle

r is the radius of the circle

Case N $1$

$x^{2}+y^{2}-2x+ 2y- 1= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-2x)+ (y^{2}+ 2y)=1$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}- 2x+1)+ (y^{2}+ 2y+1)=1+1+1$

Rewrite as perfect squares

$(x-1)^{2}+(y+1)^{2}=3$

$(x-1)^{2}+ (y+1)^{2}=\sqrt{3}^{2}$

Case N $2$

$x^{2}+ y^{2}-4x + 4y- 10= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} - 4x)+ (y^{2}+ 4y)=10$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2} - 4x+4)+ (y^{2}+ 4y+4)=10+4+4$

Rewrite as perfect squares

$(x-2)^{2}+ (y+2)^{2}=18$

$(x-2)^{2}+ (y+2)^{2}=\sqrt{18}^{2}$

Case N $3$

$x^{2}+ y^{2}-8x - 6y-20= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}- 8x)+ (y^{2} - 6y)=20$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

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

Rewrite as perfect squares

$(x-4)^{2}+ (y-3)^{2}=45$

$(x-4)^{2}+ (y-3)^{2}=\sqrt{45}^{2}$

Case N $4$  

$4x^{2}+4y^{2}+16x +24y- 40= 0$

Simplify divide by $4$ both sides

$x^{2}+ y^{2}+4x+6y- 10= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} +4x)+ (y^{2} + 6y)=10$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

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

Rewrite as perfect squares

$(x+2)^{2}+ (y+3)^{2}=23$

$(x+2)^{2}+ (y+3)^{2}=\sqrt{23}^{2}$

Case N $5$

$5x^{2}+ 5y^{2}-20x +30y+ 40= 0$

Simplify divide by $5$ both sides

$x^{2}+ y^{2}-4x +6y + 8= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} -4x)+ (y^{2}+ 6y)=-8$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-4x+4)+ (y^{2}+ 6y+9)=-8+4+9$

Rewrite as perfect squares

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

$(x-2)^{2} + (y+3)^{2}=\sqrt{5}^{2}$

Case N $6$

$2x^{2} + 2y^{2}-28x -32y-8= 0$

Simplify divide by $2$ both sides

$x^{2} + y^{2} -14x-16y-4= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} -14x)+ (y^{2} -16y)=4$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2} -14x+49)+ (y^{2} -16y+64)=4+49+64$

Rewrite as perfect squares

$(x-7)^{2} + (y-8)^{2}=117$

$(x-7)^{2} + (y-8)^{2}=\sqrt{117}^{2}$

Case N $7$

$x^{2}+ y^{2}+12x - 2y- 9 = 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} +12x)+ (y^{2} - 2y)=9$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

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

Rewrite as perfect squares

$(x+6)^{2} + (y-1)^{2}=46$      

$(x+6)^{2} + (y-1)^{2}=\sqrt{46}^{2}$

the circles in ascending order of their radius lengths is

N $1$

$x^{2}+ y^{2}-2x + 2y- 1=0$

$(x-1)^{2}+ (y+1)^{2}=\sqrt{3}^{2}$

N $2$

$5x^{2}+ 5y^{2}-20x +30y+40=0$

$(x-2)^{2}+(y+3)^{2}=\sqrt{5}^{2}$

N $3$

$x^{2}+y^{2}-4x+4y- 10=0$

$(x-2)^{2}+(y+2)^{2}=\sqrt{18}^{2}$

N $4$

$4x^{2}+4y^{2}+16x+24y-40=0$

$(x+2)^{2}+(y+3)^{2}=\sqrt{23}^{2}$

N $5$

$x^{2}+y^{2}-8x- 6y-20= 0$

$(x-4)^{2}+(y-3)^{2}=\sqrt{45}^{2}$

N $6$

$x^{2}+ y^{2}+12x- 2y-9= 0$

$(x+6)^{2}+(y-1)^{2}=\sqrt{46}^{2}$

N $7$

$2x^{2}+2y^{2}-28x-32y-8=0$

$(x-7)^{2}+(y-8)^{2}=\sqrt{117}^{2}$

Answer 2

Answer:

x2 + y2 − 2x + 2y − 1 = 0

↓

5x2 + 5y2 − 20x + 30y + 40 = 0

↓

x2 + y2 − 4x + 4y − 10 = 0

↓

4x2 + 4y2 + 16x + 24y − 40 = 0

↓

x2 + y2 − 8x − 6y − 20 = 0

↓

x2 + y2 + 12x − 2y − 9 = 0

↓

2x2 + 2y2 − 28x − 32y − 8 = 0

You're welcome again my comrades