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 Sequence

Answer 1

When doing this, we would have to first straight them all out, and as we see above, they are just all over the place, and then, we would have to set them up as a sequence.

$\left[\begin{array}{ccc}5x2 + 5y2 - 20x + 30y + 40 = 0\end{array}\right]$ would be our first step in this problem mainly because it contains the most terms in this aspect.

Then we would then $x2 + y2 - 2x + 2y -1 = 0$, then, $2x2 + 2y2 - 28x - 32y - 8 = 0$.

These would only be our first 3 part of the sequence in this aspect.

The others would then be the following:

$\boxed{x2 + y2 + 12x - 2y - 9 = 0 } \ then \ , \boxed{x2 + y2 - 4x + 4y - 10 = 0} \\ \\ then \ \boxed{x2 + y2 - 8x - 6y -20 = 0}$

Thus, as we would have one more afterward, our last part of the sequence would then be the following.

$\boxed{\boxed{4x2 + 4y2 + 16x + 24y - 40 = 0}}$

I hope this was found helpful!

Answer 2

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 $7$

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

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