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
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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 thatThe 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 radiusRemember thatConcentric circles, are circles that have the same centersoConvert each equation in standard form and then compare the centersThe complete answer in the attached documentPart 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]thereforeThe 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]thereforeThe 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]thereforeThe 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]thereforeThe 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]thereforeThe 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]thereforeThe 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]thereforeThe center is the point (-2,1) Part 8) Part 9) and Part 10) in the attached document