Answer:
The concentric circles are
[tex]3x^{2}+3y^{2}+12x-6y-21=0[/tex] and [tex]4x^{2}+4y^{2}+16x-8y-308=0[/tex]
[tex]5x^{2}+5y^{2}-30x+20y-10=0[/tex] and [tex]3x^{2}+3y^{2}-18x+12y-81=0[/tex]
[tex]4x^{2}+4y^{2}-16x+24y-28=0[/tex] and [tex]2x^{2}+2y^{2}-8x+12y-40=0[/tex]
[tex]x^{2}+y^{2}-2x+8y-13=0[/tex] and [tex]5x^{2}+5y^{2}-10x+40y-75=0[/tex]
Step-by-step explanation:
we know that
The equation of the circle in standard form is equal to
[tex](x-h)^{2} +(y-k)^{2} =r^{2}[/tex]
where
(h,k) is the center and r is the radius
Remember that
Concentric circles, are circles that have the same center
so
Convert each equation in standard form and then compare the centers
The complete answer in the attached document
Part 1) we have
[tex]3x^{2}+3y^{2}+12x-6y-21=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](3x^{2}+12x)+(3y^{2}-6y)=21[/tex]
Factor the leading coefficient of each expression
[tex]3(x^{2}+4x)+3(y^{2}-2y)=21[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]3(x^{2}+4x+4)+3(y^{2}-2y+1)=21+12+3[/tex]
[tex]3(x^{2}+4x+4)+3(y^{2}-2y+1)=36[/tex]
Rewrite as perfect squares
[tex]3(x+2)^{2}+3(y-1)^{2}=36[/tex]
[tex](x+2)^{2}+(y-1)^{2}=12[/tex]
therefore
The center is the point (-2,1)
Part 2) we have
[tex]5x^{2}+5y^{2}-30x+20y-10=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](5x^{2}-30x)+(5y^{2}+20y)=10[/tex]
Factor the leading coefficient of each expression
[tex]5(x^{2}-6x)+5(y^{2}+4y)=10[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]5(x^{2}-6x+9)+5(y^{2}+4y+4)=10+45+20[/tex]
[tex]5(x^{2}-6x+9)+5(y^{2}+4y+4)=75[/tex]
Rewrite as perfect squares
[tex]5(x-3)^{2}+5(y+2)^{2}=75[/tex]
[tex](x-3)^{2}+(y+2)^{2}=15[/tex]
therefore
The center is the point (3,-2)
Part 3) we have
[tex]x^{2}+y^{2}-12x-8y-100=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}-8y)=100[/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}-8y+16)=100+36+16[/tex]
[tex](x^{2}-12x+36)+(y^{2}-8y+16)=152[/tex]
Rewrite as perfect squares
[tex](x-6)^{2}+(y-4)^{2}=152[/tex]
therefore
The center is the point (6,4)
Part 4) we have
[tex]4x^{2}+4y^{2}-16x+24y-28=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](4x^{2}-16x)+(4y^{2}+24y)=28[/tex]
Factor the leading coefficient of each expression
[tex]4(x^{2}-4x)+4(y^{2}+6y)=28[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]4(x^{2}-4x+4)+4(y^{2}+6y+9)=28+16+36[/tex]
[tex]4(x^{2}-4x+4)+4(y^{2}+6y+9)=80[/tex]
Rewrite as perfect squares
[tex]4(x-2)^{2}+4(y+3)^{2}=80[/tex]
[tex](x-2)^{2}+(y+3)^{2}=20[/tex]
therefore
The center is the point (2,-3)
Part 5) we have
[tex]x^{2}+y^{2}-2x+8y-13=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}+8y)=13[/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}+8y+16)=13+1+16[/tex]
[tex](x^{2}-2x+1)+(y^{2}+8y+16)=30[/tex]
Rewrite as perfect squares
[tex](x-1)^{2}+(y+4)^{2}=30[/tex]
therefore
The center is the point (1,-4)
Part 6) we have
[tex]5x^{2}+5y^{2}-10x+40y-75=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](5x^{2}-10x)+(5y^{2}+40y)=75[/tex]
Factor the leading coefficient of each expression
[tex]5(x^{2}-2x)+5(y^{2}+8y)=75[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]5(x^{2}-2x+1)+5(y^{2}+8y+16)=75+5+80[/tex]
[tex]5(x^{2}-2x+1)+5(y^{2}+8y+16)=160[/tex]
Rewrite as perfect squares
[tex]5(x-1)^{2}+5(y+4)^{2}=160[/tex]
[tex](x-1)^{2}+(y+4)^{2}=32[/tex]
therefore
The center is the point (1,-4)
Part 7) we have
[tex]4x^{2}+4y^{2}+16x-8y-308=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](4x^{2}+16x)+(4y^{2}-8y)=308[/tex]
Factor the leading coefficient of each expression
[tex]4(x^{2}+4x)+4(y^{2}-2y)=308[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex]4(x^{2}+4x+4)+4(y^{2}-2y+1)=308+16+4[/tex]
[tex]4(x^{2}+4x+4)+4(y^{2}-2y+1)=328[/tex]
Rewrite as perfect squares
[tex]4(x+2)^{2}+4(y-1)^{2}=328[/tex]
[tex](x+2)^{2}+(y-1)^{2}=82[/tex]
therefore
The center is the point (-2,1)
Part 8) Part 9) and Part 10) in the attached document
