Question

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. Pleaseeeeeeee help

Answer 1

Answer:

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

so

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

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