MATH SOLVE

3 months ago

Q:
# 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

Accepted Solution

A:

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.

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

Then we would then [tex]x2 + y2 - 2x + 2y -1 = 0[/tex], then, [tex]2x2 + 2y2 - 28x - 32y - 8 = 0[/tex].

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

The others would then be the following:

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

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

[tex]\boxed{\boxed{4x2 + 4y2 + 16x + 24y - 40 = 0}}[/tex]

I hope this was found helpful!

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

Then we would then [tex]x2 + y2 - 2x + 2y -1 = 0[/tex], then, [tex]2x2 + 2y2 - 28x - 32y - 8 = 0[/tex].

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

The others would then be the following:

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

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

[tex]\boxed{\boxed{4x2 + 4y2 + 16x + 24y - 40 = 0}}[/tex]

I hope this was found helpful!