Question

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.

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

Find the radius of each equation:

1.

`x^2 + y^2-2x+2y-1 = 0, \\ x^2-2x+1-1 + y^2+2y+1-1-1 = 0, \\ (x-1)^2+(y+1)^2=3`, then
`r_1= √(3)`.

2.

`x^2 + y^2-4x + 4y- 10 = 0, \\ x^2 -4x+4-4+ y^2 + 4y+4-4- 10 = 0, \\ (x-2)^2+(y+2)^2=18`, then
`r_2= √(18)=3 √(2)`.

3.

`x^2 + y^2-8x- 6y- 20 = 0, \\ x^2-8x+16-16+ y^2- 6y+9-9- 20 = 0, \\ (x-4)^2+(y-3)^2=45`, then
`r_3= √(45) =3 √(5)`.


4.

`4x^2 + 4y^2+16x+24y- 40 = 0, \\ 4x^2+16x+16-16+ 4y^2+24y+36-36- 40 = 0, \\ 4(x+2)^2+4(y+3)^2=92,\\ (x+2)^2+(y+3)^2=23`, then
`r_4= √(23)`.

5.

`5x^2 + 5y^2-20x+30y+ 40 = 0, \\ 5x^2-20x+20-20+ 5y^2+30y+45-45- 40 = 0, \\ 5(x-2)^2+5(y+3)^2=105,\\ (x-2)^2+(y+3)^2=21`, then
`r_5= √(21)`.

6.

`2x^2 + 2y^2-28x-32y- 8= 0, \\ 2x^2-28x+98-98+ 2y^2-32y+128-128- 8= 0, \\ 2(x-7)^2+2(y-8)^2=234,\\ (x+2)^2+(y+3)^2=117`, then
`r_6= √(117)=3√(13)`.

7.

`x^2 + y^2+12x-2y-9 = 0, \\ x^2+12x+36-36+ y^2-2y+1-1- 9 = 0, \\ (x+6)^2+(y-1)^2=46`, then
`r_7= √(46)`.

Hence

`r_1= √(3)`,
`r_2=3 √(2)`,
`r_3=3 √(5)`,
`r_4= √(23)`,
`r_5= √(21)`,
`r_6= 3√(13)`,
`r_7= √(46)` and
`r_1\ \textless \ r_2\ \textless \ r_5\ \textless \ r_4\ \textless \ r_3\ \textless \ r_7\ \textless \ r_6`.

Answer 2

The correct answer is:

x²+y²-2x+2y-1 = 0;
x²+y²-4x+4y-10 = 0;
5x²+5y²-20x+30y+40 = 0;
x²+y²-8x-6y-20 = 0;
x²+y²+12x-2y-9 = 0;
4x²+4y²+16x+24y-40 = 0; and
2x²+2y²-28x-32y-8 = 0

Step-by-step explanation:

For each of these, we want to write the equation in the form
(x+h)²+(y+k)² = r².

To do this, we evaluate the terms 2hx and 2ky in each equation. We will take half of this; this will tell us what h and k are for each equation.

For the first equation:
2hx = -2x and 2ky = 2y.

Half of -2x = -1x and half of 2y = 1y; this means h = -1 and k = 1:
(x-1)² + (y+1)² + ___ - 1 = 0

When we multiply (x-1)², we get
x²-2x+1.
When we multiply (y+1)², we get
y²+2y+1.

This gives us 1+1 = 2 for the constant. We know we must add something to 2 to get -1; 2 + ___ = -1; the missing term is -3. Add that to each side (to have r² on the right side of the equals) and we have
(x-1)² + (y+1)² = 3
This means that r² = 3, and r = √3 = 1.732.

For the second equation, 2hx = -4x and 2ky = 4y; this means h = -4/2 = -2 and k = 4/2 = 2. This gives us
(x-2)² + (y+2)² -10 + ___ = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+2)² gives us
y²+4x+4.
This gives us 4+4= 8 for our constant so far.

We know 8 + ___ = -10; this means the missing term is -18. Add this to each side of the equation to have
(x-2)²+(y+2)² = 18; r² = 18; r = √18 = 3√2 = 4.243.

For the third equation, 2hx = -8x and 2ky = -6y. This means h = -8/2 = -4 and k = -6/2 = -3. This gives us:
(x-4)²+(y-3)²-20 = 0

Multiplying (x-4)² gives us
x²-8x+16.
Multiplying (y-3)² gives us
y²-6y+9.

This gives us 16+9 = 25 for the constant. We know that 25+___ = -20; the missing term is -45. Add this to each side for r², and we have that
r²=45; r = √45 = 3√5 = 6.708.

For the next equation, we factor 4 out of the entire equation:
4(x²+y²+4x+6y-10)=0.
This means 2hx = 4x and 2ky = 6y; this gives us h = 4/2 = 2 and k = 6/2 = 3. This gives us
4((x+2)²+(y+3)² - 10) = 0.

Multiplying (x+2)² gives us
x²+4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know 13+__ = -10; this missing value is -23. Since we had factored out a 4, that means we have 4(-23) = -92. Adding this to each side for r², we have
r²=92; r = √92 = 2√23 = 9.59.

For the next equation, we factor out a 5 first:
5(x²+y²-4x+6y+8) = 0. This means that 2hx = -4x and 2ky = 6y; this gives us h = -4/2 = -2 and k = 6/2 = 3:

5((x-2)²+(y+3)²+8) = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know that 13+__ = 8; the missing value is -5. Since we factored a 5 out, we have 5(-5) = -25. Adding this to each side for r² gives us
r²=25; r = √25 = 5.

For the next equation, we first factor a 2 out:
2(x²+y²-14x-16y-4) = 0. This means 2hx = -14x and 2ky = -16y; this gives us h = -14/2 = -7 and k = -16/2 = -8:

2((x-7)²+(y-8)²-4) = 0.

Multiplying (x-7)² gives us
x²-14x+49.
Multiplying (y-8)² gives us
y²-16x+64.

This gives us a constant of 49+64=113. We know that 113+__ = -4; the missing value is -117. Since we first factored out a 2, this gives us 2(-117) = -234. Adding this to each side for r² gives us
r²=234; r = √234 = 3√26 = 15.297.

For the last equation, 2hx = 12x and 2ky = -2; this means h = 12/2 = 6 and k = -2/2 = -1:
(x+6)²+(y-1)²-9 = 0

Multiplying (x+6)² gives us
x²+12x+36.
Multiplying (y-1)² gives us
y²-2y+1.

This gives us a constant of 36+1 = 37. We know that 37+__ = -9; the missing value is -46. Adding this to each side for r² gives us
r² = 46; r=√46 = 6.78.