Mathematics, 22.07.2019 01:10 bossboybaker

Drag the titles to the boxes to form correct pairs .not all titles will be used. match the pairs of equation that represents concentric circles.

Drag the titles to the boxes to form correct pairs .not all titles will be used. match the pairs of

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Answer from: jourdyno

The concentric circles are

$3x^{2}+3y^{2}+12x-6y-21=0$ and $4x^{2}+4y^{2}+16x-8y-308=0$

$5x^{2}+5y^{2}-30x+20y-10=0$ and $3x^{2}+3y^{2}-18x+12y-81=0$

$4x^{2}+4y^{2}-16x+24y-28=0$ and $2x^{2}+2y^{2}-8x+12y-40=0$

$x^{2}+y^{2}-2x+8y-13=0$ and $5x^{2}+5y^{2}-10x+40y-75=0$

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2} =r^{2}$

where

(h,k) is the center and r is the radius

Remember that

Concentric circles, are circles that have the same center

Convert each equation in standard form and then compare the centers

The complete answer in the attached document

Part 1) we have

$3x^{2}+3y^{2}+12x-6y-21=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(3x^{2}+12x)+(3y^{2}-6y)=21$

Factor the leading coefficient of each expression

$3(x^{2}+4x)+3(y^{2}-2y)=21$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$3(x^{2}+4x+4)+3(y^{2}-2y+1)=21+12+3$

$3(x^{2}+4x+4)+3(y^{2}-2y+1)=36$

Rewrite as perfect squares

$3(x+2)^{2}+3(y-1)^{2}=36$

$(x+2)^{2}+(y-1)^{2}=12$

therefore

The center is the point (-2,1)

Part 2) we have

$5x^{2}+5y^{2}-30x+20y-10=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(5x^{2}-30x)+(5y^{2}+20y)=10$

Factor the leading coefficient of each expression

$5(x^{2}-6x)+5(y^{2}+4y)=10$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$5(x^{2}-6x+9)+5(y^{2}+4y+4)=10+45+20$

$5(x^{2}-6x+9)+5(y^{2}+4y+4)=75$

Rewrite as perfect squares

$5(x-3)^{2}+5(y+2)^{2}=75$

$(x-3)^{2}+(y+2)^{2}=15$

therefore

The center is the point (3,-2)

Part 3) we have

$x^{2}+y^{2}-12x-8y-100=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-12x)+(y^{2}-8y)=100$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-12x+36)+(y^{2}-8y+16)=100+36+16$

$(x^{2}-12x+36)+(y^{2}-8y+16)=152$

Rewrite as perfect squares

$(x-6)^{2}+(y-4)^{2}=152$

therefore

The center is the point (6,4)

Part 4) we have

$4x^{2}+4y^{2}-16x+24y-28=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(4x^{2}-16x)+(4y^{2}+24y)=28$

Factor the leading coefficient of each expression

$4(x^{2}-4x)+4(y^{2}+6y)=28$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$4(x^{2}-4x+4)+4(y^{2}+6y+9)=28+16+36$

$4(x^{2}-4x+4)+4(y^{2}+6y+9)=80$

Rewrite as perfect squares

$4(x-2)^{2}+4(y+3)^{2}=80$

$(x-2)^{2}+(y+3)^{2}=20$

therefore

The center is the point (2,-3)

Part 5) we have

$x^{2}+y^{2}-2x+8y-13=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-2x)+(y^{2}+8y)=13$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-2x+1)+(y^{2}+8y+16)=13+1+16$

$(x^{2}-2x+1)+(y^{2}+8y+16)=30$

Rewrite as perfect squares

$(x-1)^{2}+(y+4)^{2}=30$

therefore

The center is the point (1,-4)

Part 6) we have

$5x^{2}+5y^{2}-10x+40y-75=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(5x^{2}-10x)+(5y^{2}+40y)=75$

Factor the leading coefficient of each expression

$5(x^{2}-2x)+5(y^{2}+8y)=75$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$5(x^{2}-2x+1)+5(y^{2}+8y+16)=75+5+80$

$5(x^{2}-2x+1)+5(y^{2}+8y+16)=160$

Rewrite as perfect squares

$5(x-1)^{2}+5(y+4)^{2}=160$

$(x-1)^{2}+(y+4)^{2}=32$

therefore

The center is the point (1,-4)

Part 7) we have

$4x^{2}+4y^{2}+16x-8y-308=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(4x^{2}+16x)+(4y^{2}-8y)=308$

Factor the leading coefficient of each expression

$4(x^{2}+4x)+4(y^{2}-2y)=308$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$4(x^{2}+4x+4)+4(y^{2}-2y+1)=308+16+4$

$4(x^{2}+4x+4)+4(y^{2}-2y+1)=328$

Rewrite as perfect squares

$4(x+2)^{2}+4(y-1)^{2}=328$

$(x+2)^{2}+(y-1)^{2}=82$

therefore

The center is the point (-2,1)

Part 8) Part 9) and Part 10) in the attached document

Answer from: Quest

False because how does that work?

Answer from: Quest

i don't even know man. edge inuity be messing with us. i had to ask a question too.

You can use a system of equations to graph and solve the polynomial equation 3x3 + x=2x2 +. which st

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Drag the titles to the boxes to form correct pairs .not all titles will be used. match the pairs of equation that represents concentric circles.

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