Respuesta :
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
[tex](x-h)^{2} +(y-k)^{2} =r^{2}[/tex]
where
(h,k) is the center of the circle
r is the radius of the circle
Case N [tex]1[/tex]
[tex]x^{2}+y^{2}-2x+ 2y- 1= 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}-2x)+ (y^{2}+ 2y)=1[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}- 2x+1)+ (y^{2}+ 2y+1)=1+1+1[/tex]
Rewrite as perfect squares
[tex](x-1)^{2}+(y+1)^{2}=3[/tex]
[tex](x-1)^{2}+ (y+1)^{2}=\sqrt{3}^{2} [/tex]
Case N [tex]2[/tex]
[tex]x^{2}+ y^{2}-4x + 4y- 10= 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2} - 4x)+ (y^{2}+ 4y)=10[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2} - 4x+4)+ (y^{2}+ 4y+4)=10+4+4[/tex]
Rewrite as perfect squares
[tex](x-2)^{2}+ (y+2)^{2}=18[/tex]
[tex](x-2)^{2}+ (y+2)^{2}=\sqrt{18}^{2}[/tex]
Case N [tex]3[/tex]
[tex]x^{2}+ y^{2}-8x - 6y-20= 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}- 8x)+ (y^{2} - 6y)=20[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}- 8x+16)+ (y^{2}- 6y+9)=20+16+9[/tex]
Rewrite as perfect squares
[tex](x-4)^{2}+ (y-3)^{2}=45[/tex]
[tex](x-4)^{2}+ (y-3)^{2}=\sqrt{45}^{2}[/tex]
Case N [tex]4[/tex]
[tex]4x^{2}+4y^{2}+16x +24y- 40= 0[/tex]
Simplify divide by [tex]4[/tex] both sides
[tex]x^{2}+ y^{2}+4x+6y- 10= 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2} +4x)+ (y^{2} + 6y)=10[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}+4x+4)+(y^{2} + 6y+9)=10+4+9[/tex]
Rewrite as perfect squares
[tex](x+2)^{2}+ (y+3)^{2}=23[/tex]
[tex](x+2)^{2}+ (y+3)^{2}=\sqrt{23}^{2}[/tex]
Case N [tex]5[/tex]
[tex]5x^{2}+ 5y^{2}-20x +30y+ 40= 0[/tex]
Simplify divide by [tex]5[/tex] both sides
[tex]x^{2}+ y^{2}-4x +6y + 8= 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2} -4x)+ (y^{2}+ 6y)=-8[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}-4x+4)+ (y^{2}+ 6y+9)=-8+4+9[/tex]
Rewrite as perfect squares
[tex](x-2)^{2} + (y+3)^{2}=5[/tex]
[tex](x-2)^{2} + (y+3)^{2}=\sqrt{5}^{2}[/tex]
Case N [tex]6[/tex]
[tex]2x^{2} + 2y^{2}-28x -32y-8= 0[/tex]
Simplify divide by [tex]2[/tex] both sides
[tex]x^{2} + y^{2} -14x-16y-4= 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2} -14x)+ (y^{2} -16y)=4[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2} -14x+49)+ (y^{2} -16y+64)=4+49+64[/tex]
Rewrite as perfect squares
[tex](x-7)^{2} + (y-8)^{2}=117[/tex]
[tex](x-7)^{2} + (y-8)^{2}=\sqrt{117}^{2}[/tex]
Case N [tex]7[/tex]
[tex]x^{2}+ y^{2}+12x - 2y- 9 = 0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2} +12x)+ (y^{2} - 2y)=9[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2} +12x+36)+ (y^{2} - 2y+1)=9+36+1[/tex]
Rewrite as perfect squares
[tex](x+6)^{2} + (y-1)^{2}=46[/tex]
[tex](x+6)^{2} + (y-1)^{2}=\sqrt{46}^{2}[/tex]
the circles in ascending order of their radius lengths is
N [tex]1[/tex]
[tex]x^{2}+ y^{2}-2x + 2y- 1=0[/tex]
[tex](x-1)^{2}+ (y+1)^{2}=\sqrt{3}^{2} [/tex]
N [tex]2[/tex]
[tex]5x^{2}+ 5y^{2}-20x +30y+40=0[/tex]
[tex](x-2)^{2}+(y+3)^{2}=\sqrt{5}^{2}[/tex]
N [tex]3[/tex]
[tex]x^{2}+y^{2}-4x+4y- 10=0[/tex]
[tex](x-2)^{2}+(y+2)^{2}=\sqrt{18}^{2}[/tex]
N [tex]4[/tex]
[tex]4x^{2}+4y^{2}+16x+24y-40=0[/tex]
[tex](x+2)^{2}+(y+3)^{2}=\sqrt{23}^{2}[/tex]
N [tex]5[/tex]
[tex]x^{2}+y^{2}-8x- 6y-20= 0[/tex]
[tex](x-4)^{2}+(y-3)^{2}=\sqrt{45}^{2}[/tex]
N [tex]6[/tex]
[tex]x^{2}+ y^{2}+12x- 2y-9= 0[/tex]
[tex](x+6)^{2}+(y-1)^{2}=\sqrt{46}^{2}[/tex]
N [tex]7[/tex]
[tex]2x^{2}+2y^{2}-28x-32y-8=0[/tex]
[tex](x-7)^{2}+(y-8)^{2}=\sqrt{117}^{2}[/tex]
The given equations can be rearranged in the standard form of the
equation of a circle, from which the radius can be evaluated.
Response:
The equations arranged in increasing order of their radius lengths are;
x² + y² - 2·x + 2·y - 1 = 0
[tex]{}[/tex]↓
5·x² + 5·y² - 20·x + 30·y + 40 = 0
[tex]{}[/tex]↓
x² + y² - 4·x + 4·y - 10 = 0
[tex]{}[/tex]↓
4·x² + 4·y² + 16·x + 24·y - 40 = 0
[tex]{}[/tex]↓
2·x² + 2·y² - 28·x - 32·y - 8 = 0
[tex]{}[/tex]↓
x² + y² - 8·x - 6·y - 20 = 0
[tex]{}[/tex]↓
x² + y² + 12·x - 2·y - 9 = 0
[tex]{}[/tex]↓
2·x² + 2·y² - 28·x - 32·y - 8 = 0
How can the general forms of the given equations be arranged in standard form?
The given equation are;
- x² + y² - 2·x + 2·y - 1 = 0
- x² + y² - 4·x + 4·y - 10 = 0
- x² + y² - 8·x - 6·y - 20 = 0
- 4·x² + 4·y² + 16·x + 24·y - 40 = 0
- 5·x² + 5·y² - 20·x + 30·y + 40 = 0
- 2·x² + 2·y² - 28·x - 32·y - 8 = 0
- x² + y² + 12·x - 2·y - 9 = 0
The standard form of he equation of a circle is;
- (x - h)² + (y - k)² = r²
Where;
r = The radius length
- x² + y² - 2·x + 2·y - 1 = 0
x² - 2·x + 1 + 3y² + 2·y + 1 = 1 + 1 + 1 = 3
(x - 1)² + (y + 1)² = 3
Therefore;
r² = 3
- r = √3
x² + y² - 4·x + 4·y - 10 = 0
x² - 4·x + y²+ 4·y - 10 = 0
x² - 4·x + 4 + y²+ 4·y + 4 = 10 + 4 + 4 = 18
(x - 2)² + (y + 2)² = 18
r² = 18
- r = 3·√2
x² + y² - 8·x - 6·y - 20 = 0
x² - 8·x + y² - 6·y - 20 = 0
x² - 8·x + y² - 6·y = 20
x² - 8·x + 16 + y² - 6·y + 9 = 20 + 16 + 9 = 45
(x - 4)² + (y - 3)² = 45
- r = √45
4·x² + 4·y² + 16·x + 24·y - 40 = 0
4·(x² + 4·x + y² + 6·y - 10) = 0
x² + 4·x + y² + 6·y - 10 = 0
x² + 4·x + y² + 6·y = 10
x² + 4·x + 4 + y² + 6·y + 9 = 10 + 4 + 9
(x + 2)² + (y + 3)² = 23
- r = √23
5·x² + 5·y² - 20·x + 30·y + 40 = 0
5·(x² + y² - 4·x + 6·y + 8) = 0
x² + y² - 4·x + 6·y + 8 = 0
x² - 4·x + y² - 6·y + 8 = 0
x² - 4·x + 4 + y² - 6·y + 9 = -8
(x - 2)² + (y - 3)² = -8 + 4 + 9 = 5
- r = √(5)
2·x² + 2·y² - 28·x - 32·y - 8 = 0
2·(x² + y² - 14·x - 16·y - 4) = 0
x² + y² - 14·x - 16·y - 4 = 0
x² - 14·x + y²- 16·y = 4
x² - 14·x + 49 + y²- 16·y + 64 = 4 + 49 + 64 = 117
(x - 7)² + (y - 8)² = 117
- r = √(117) = 3·√(13)
x² + y² + 12·x - 2·y - 9 = 0
x² + 12·x + y² - 2·y - 9 = 0
x² + 12·x + y² - 2·y = 9
x² + 12·x + 36 + y² - 2·y + 1 = 9 + 36 + 1
(x + 6)² + (y - 1)² = 46
- r = √46
The radius in ascending order are therefore;
r = √(3) → r = √5 → r = 3·√2 → r = √23 → r = √45 → r = √46 → r = 3·√(13)
Which gives;
x² + y² - 2·x + 2·y - 1 = 0
[tex]{}[/tex]↓
5·x² + 5·y² - 20·x + 30·y + 40 = 0
[tex]{}[/tex]↓
x² + y² - 4·x + 4·y - 10 = 0
[tex]{}[/tex]↓
4·x² + 4·y² + 16·x + 24·y - 40 = 0
[tex]{}[/tex]↓
2·x² + 2·y² - 28·x - 32·y - 8 = 0
[tex]{}[/tex]↓
x² + y² - 8·x - 6·y - 20 = 0
[tex]{}[/tex]↓
x² + y² + 12·x - 2·y - 9 = 0
[tex]{}[/tex]↓
2·x² + 2·y² - 28·x - 32·y - 8 = 0
Learn more about the standard equation of a circle here:
https://brainly.com/question/502872
