Mathematics, 18.07.2019 11:50 mooreadrian412

Match the circle equations in general form with their corresponding equations in standard form. tilesx2 + y2− 4x + 12y − 20 = 0(x − 6)2+ (y − 4)2 = 56x2 + y2+ 6x − 8y − 10 = 0 (x − 2)2+ (y + 6)2 = 603x2 + 3y2 + 12x + 18y − 15 = 0(x + 2)2+ (y + 3)2 = 185x2 + 5y2 − 10x + 20y − 30 = 0(x + 1)2+ (y − 6)2 = 462x2 + 2y2 − 24x − 16y − 8 = 0x2 + y2+ 2x − 12y − 9 =

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Answer from: oreo543

case A) $x^{2} +y^{2} -4x+12y-20=0$

Convert to standard form

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

$(x^{2}-4x)+(y^{2}+12y)=20$

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

$(x^{2}-4x+4)+(y^{2}+12y+36)=20+4+36$

$(x^{2}-4x+4)+(y^{2}+12y+36)=60$

Rewrite as perfect squares

$(x-2)^{2}+(y+6)^{2}=60$

therefore

the answer case A) is

$x^{2} +y^{2} -4x+12y-20=0$ -----> $(x-2)^{2}+(y+6)^{2}=60$

case B) $x^{2} +y^{2} +6x-8y-10=0$

Convert to standard form

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

$(x^{2}+6x)+(y^{2}-8y)=10$

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

$(x^{2}+6x+9)+(y^{2}-8y+16)=10+9+16$

$(x^{2}+6x+9)+(y^{2}-8y+16)=35$

Rewrite as perfect squares

$(x+3)^{2}+(y-4)^{2}=35$

therefore

the answer case B) is

$x^{2} +y^{2} +6x-8y-10=0$ -----> $(x+3)^{2}+(y-4)^{2}=35$

case C) $3x^{2} +3y^{2} +12x+18y-15=0$

Convert to standard form

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

$(3x^{2}+12x)+(3y^{2}+18y)=15$

Factor the leading coefficient of each expression

$3(x^{2}+4x)+3(y^{2}+6y)=15$

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

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

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

$(x^{2}+4x+4)+(y^{2}+6y+9)=18$

Rewrite as perfect squares

$(x+2)^{2}+(y+3)^{2}=18$

therefore

the answer case C) is

$3x^{2} +3y^{2} +12x+18y-15=0$ -----> $(x+2)^{2}+(y+3)^{2}=18$

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

Convert to standard form

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

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

Factor the leading coefficient of each expression

$5(x^{2}-2x)+5(y^{2}+4y)=30$

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

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

$(x^{2}-2x+1)+(y^{2}+4y+4)=6+1+4$

$(x^{2}-2x+1)+(y^{2}+4y+4)=11$

Rewrite as perfect squares

$(x-1)^{2}+(y+2)^{2}=11$

therefore

the answer case D) is

$3x^{2} +3y^{2} +12x+18y-15=0$ -----> $(x-1)^{2}+(y+2)^{2}=11$

case E) $2x^{2} +2y^{2} -24x-16y-8=0$

Convert to standard form

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

$(2x^{2}-24x)+(2y^{2}-16y)=8$

Factor the leading coefficient of each expression

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

$(x^{2}-12x)+(y^{2}-8y)=4$

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

$(x^{2}-12x+36)+(y^{2}-8y+16)=4+36+16$

$(x^{2}-12x+36)+(y^{2}-8y+16)=56$

Rewrite as perfect squares

$(x-6)^{2}+(y-4)^{2}=56$

therefore

the answer case E) is

$2x^{2} +2y^{2} -24x-16y-8=0$ -----> $(x-6)^{2}+(y-4)^{2}=56$

case F) $x^{2} +y^{2}+2x-12y-9=0$

Convert to standard form

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

$(x^{2}+2x)+(y^{2}-12y)=9$

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

$(x^{2}+2x+1)+(y^{2}-12y+36)=9+1+36$

$(x^{2}+2x+1)+(y^{2}-12y+36)=46$

Rewrite as perfect squares

$(x+1)^{2}+(y-6)^{2}=46$

therefore

the answer case F) is

$x^{2} +y^{2}+2x-12y-9=0$ -----> $(x+1)^{2}+(y-6)^{2}=46$

Answer from: Quest

asdasd

step-by-step explanation:

Answer from: Quest

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2 is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes. our only condition is that it passes through the point (-2,5)

A. write an equation of a line that passes through (-2, 5). b. how many different equations could y

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Match the circle equations in general form with their corresponding equations in standard form. tile...

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