Answer :
To determine which pairs of given equations represent concentric circles, we need to find the centers of each circle from the given equations.
The general form of the circle equation is:
[tex]\[ Ax^2 + Ay^2 + Dx + Ey + F = 0 \][/tex]
We can convert this into standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius, by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
Let's work on each equation:
1. [tex]\(3x^2 + 3y^2 + 12x - 6y - 21 = 0\)[/tex]
[tex]\[ x^2 + y^2 + 4x - 2y = 7 \\ (x^2 + 4x + 4) + (y^2 - 2y + 1) = 7 + 4 + 1 \\ (x+2)^2 + (y-1)^2 = 12 \\ \text{Center: } (-2, 1) \][/tex]
2. [tex]\(5x^2 + 5y^2 - 10x + 40y - 75 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 2x + 8y = 15 \\ (x^2 - 2x + 1) + (y^2 + 8y + 16) = 15 + 1 + 16 \\ (x-1)^2 + (y+4)^2 = 32 \\ \text{Center: } (1, -4) \][/tex]
3. [tex]\(5x^2 + 5y^2 - 30x + 20y - 10 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 6x + 4y = 2 \\ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 2 + 9 + 4 \\ (x-3)^2 + (y+2)^2 = 15 \\ \text{Center: } (3, -2) \][/tex]
4. [tex]\(4x^2 + 4y^2 + 16x - 8y - 308 = 0\)[/tex]
[tex]\[ x^2 + y^2 + 4x - 2y = 77 \\ (x^2 + 4x + 4) + (y^2 - 2y + 1) = 77 + 4 + 1 \\ (x+2)^2 + (y-1)^2 = 82 \\ \text{Center: } (-2, 1) \][/tex]
5. [tex]\(x^2 + y^2 - 12x - 8y - 100 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 12x - 8y = 100 \\ (x^2 - 12x + 36) + (y^2 - 8y + 16) = 100 + 36 + 16 \\ (x-6)^2 + (y-4)^2 = 152 \\ \text{Center: } (6, 4) \][/tex]
6. [tex]\(2x^2 + 2y^2 - 8x + 12y - 40 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 4x + 6y = 20 \\ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 20 + 4 + 9 \\ (x-2)^2 + (y+3)^2 = 33 \\ \text{Center: } (2, -3) \][/tex]
7. [tex]\(4x^2 + 4y^2 - 16x + 24y - 28 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 4x + 6y = 7 \\ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 7 + 4 + 9 \\ (x-2)^2 + (y+3)^2 = 20 \\ \text{Center: } (2, -3) \][/tex]
8. [tex]\(3x^2 + 3y^2 - 18x + 12y - 81 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 6x + 4y = 27 \\ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 27 + 9 + 4 \\ (x-3)^2 + (y+2)^2 = 40 \\ \text{Center: } (3, -2) \][/tex]
9. [tex]\(x^2 + y^2 - 2x + 8y - 13 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 2x + 8y = 13 \\ (x^2 - 2x + 1) + (y^2 + 8y + 16) = 13 + 1 + 16 \\ (x-1)^2 + (y+4)^2 = 30 \\ \text{Center: } (1, -4) \][/tex]
10. [tex]\(x^2 + y^2 + 24x + 30y + 17 = 0\)[/tex]
[tex]\[ x^2 + y^2 + 24x + 30y = -17 \\ (x + 12)^2 + (y + 15)^2 = 190 \\ \text{Center: } (-12, -15) \][/tex]
Now, we need to pair the circles with identical centers:
- Equations (1) and (4) have the same center (-2, 1).
- Equations (6) and (7) have the same center (2, -3).
- Equations (3) and (8) have the same center (3, -2).
- Equations (2) and (9) have the same center (1, -4).
Let's place them correctly:
[tex]$ \begin{array}{l} 3 x^2+3 y^2+12 x-6 y-21=0 \\ 4 x^2+4 y^2+16 x-8 y-308=0 \\ 2 x^2+2 y^2-8 x+12 y-40=0 \\ 4 x^2+4 y^2-16 x+24 y-28=0 \\ 5 x^2+5 y^2-30 x+20 y-10=0 \\ 3 x^2+3 y^2-18 x+12 y-81=0 \\ 5 x^2+5 y^2-10 x+40 y-75=0 \\ x^2+y^2-2 x+8 y-13=0 \\ \end{array} $[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\qquad$[/tex]
[tex]$\longleftrightarrow$[/tex]
[tex]$\square$[/tex]
[tex]$ \stackrel{\rightharpoonup}{\longleftrightarrow} $[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
Equations:
- [tex]\(3x^2 + 3y^2 + 12x - 6y - 21 = 0\)[/tex] and [tex]\(4x^2 + 4y^2 + 16x - 8y - 308 = 0\)[/tex]
- [tex]\(2x^2 + 2y^2 - 8x + 12y - 40 = 0\)[/tex] and [tex]\(4x^2 + 4y^2 - 16x + 24y - 28 = 0\)[/tex]
- [tex]\(5x^2 + 5y^2 - 30x + 20y - 10 = 0\)[/tex] and [tex]\(3x^2 + 3y^2 - 18x + 12y - 81 = 0\)[/tex]
- [tex]\(5x^2 + 5y^2 - 10x + 40y - 75 = 0\)[/tex] and [tex]\(x^2 + y^2 - 2x + 8y - 13 = 0\)[/tex]
The general form of the circle equation is:
[tex]\[ Ax^2 + Ay^2 + Dx + Ey + F = 0 \][/tex]
We can convert this into standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius, by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
Let's work on each equation:
1. [tex]\(3x^2 + 3y^2 + 12x - 6y - 21 = 0\)[/tex]
[tex]\[ x^2 + y^2 + 4x - 2y = 7 \\ (x^2 + 4x + 4) + (y^2 - 2y + 1) = 7 + 4 + 1 \\ (x+2)^2 + (y-1)^2 = 12 \\ \text{Center: } (-2, 1) \][/tex]
2. [tex]\(5x^2 + 5y^2 - 10x + 40y - 75 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 2x + 8y = 15 \\ (x^2 - 2x + 1) + (y^2 + 8y + 16) = 15 + 1 + 16 \\ (x-1)^2 + (y+4)^2 = 32 \\ \text{Center: } (1, -4) \][/tex]
3. [tex]\(5x^2 + 5y^2 - 30x + 20y - 10 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 6x + 4y = 2 \\ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 2 + 9 + 4 \\ (x-3)^2 + (y+2)^2 = 15 \\ \text{Center: } (3, -2) \][/tex]
4. [tex]\(4x^2 + 4y^2 + 16x - 8y - 308 = 0\)[/tex]
[tex]\[ x^2 + y^2 + 4x - 2y = 77 \\ (x^2 + 4x + 4) + (y^2 - 2y + 1) = 77 + 4 + 1 \\ (x+2)^2 + (y-1)^2 = 82 \\ \text{Center: } (-2, 1) \][/tex]
5. [tex]\(x^2 + y^2 - 12x - 8y - 100 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 12x - 8y = 100 \\ (x^2 - 12x + 36) + (y^2 - 8y + 16) = 100 + 36 + 16 \\ (x-6)^2 + (y-4)^2 = 152 \\ \text{Center: } (6, 4) \][/tex]
6. [tex]\(2x^2 + 2y^2 - 8x + 12y - 40 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 4x + 6y = 20 \\ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 20 + 4 + 9 \\ (x-2)^2 + (y+3)^2 = 33 \\ \text{Center: } (2, -3) \][/tex]
7. [tex]\(4x^2 + 4y^2 - 16x + 24y - 28 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 4x + 6y = 7 \\ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 7 + 4 + 9 \\ (x-2)^2 + (y+3)^2 = 20 \\ \text{Center: } (2, -3) \][/tex]
8. [tex]\(3x^2 + 3y^2 - 18x + 12y - 81 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 6x + 4y = 27 \\ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 27 + 9 + 4 \\ (x-3)^2 + (y+2)^2 = 40 \\ \text{Center: } (3, -2) \][/tex]
9. [tex]\(x^2 + y^2 - 2x + 8y - 13 = 0\)[/tex]
[tex]\[ x^2 + y^2 - 2x + 8y = 13 \\ (x^2 - 2x + 1) + (y^2 + 8y + 16) = 13 + 1 + 16 \\ (x-1)^2 + (y+4)^2 = 30 \\ \text{Center: } (1, -4) \][/tex]
10. [tex]\(x^2 + y^2 + 24x + 30y + 17 = 0\)[/tex]
[tex]\[ x^2 + y^2 + 24x + 30y = -17 \\ (x + 12)^2 + (y + 15)^2 = 190 \\ \text{Center: } (-12, -15) \][/tex]
Now, we need to pair the circles with identical centers:
- Equations (1) and (4) have the same center (-2, 1).
- Equations (6) and (7) have the same center (2, -3).
- Equations (3) and (8) have the same center (3, -2).
- Equations (2) and (9) have the same center (1, -4).
Let's place them correctly:
[tex]$ \begin{array}{l} 3 x^2+3 y^2+12 x-6 y-21=0 \\ 4 x^2+4 y^2+16 x-8 y-308=0 \\ 2 x^2+2 y^2-8 x+12 y-40=0 \\ 4 x^2+4 y^2-16 x+24 y-28=0 \\ 5 x^2+5 y^2-30 x+20 y-10=0 \\ 3 x^2+3 y^2-18 x+12 y-81=0 \\ 5 x^2+5 y^2-10 x+40 y-75=0 \\ x^2+y^2-2 x+8 y-13=0 \\ \end{array} $[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\qquad$[/tex]
[tex]$\longleftrightarrow$[/tex]
[tex]$\square$[/tex]
[tex]$ \stackrel{\rightharpoonup}{\longleftrightarrow} $[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
Equations:
- [tex]\(3x^2 + 3y^2 + 12x - 6y - 21 = 0\)[/tex] and [tex]\(4x^2 + 4y^2 + 16x - 8y - 308 = 0\)[/tex]
- [tex]\(2x^2 + 2y^2 - 8x + 12y - 40 = 0\)[/tex] and [tex]\(4x^2 + 4y^2 - 16x + 24y - 28 = 0\)[/tex]
- [tex]\(5x^2 + 5y^2 - 30x + 20y - 10 = 0\)[/tex] and [tex]\(3x^2 + 3y^2 - 18x + 12y - 81 = 0\)[/tex]
- [tex]\(5x^2 + 5y^2 - 10x + 40y - 75 = 0\)[/tex] and [tex]\(x^2 + y^2 - 2x + 8y - 13 = 0\)[/tex]
