Question

Find the center coordinates and radius of a circle for an equation written as: 3x2 + 3y2 - 6y = —12× + 24

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53 Answers

Move the variable to the left-hand side and change its sign
$3{x}^{2}+3{y}^{2}-6y+12x=24$

Use the commutative property to reorder the terms
$3{x}^{2}+12x+3{y}^{2}-6y=24$

Factor out $3$ from the expression
$3\left( {x}^{2}+4x \right)+3{y}^{2}-6y=24$

To complete the square, the same value needs to be added to both sides
$3\left( {x}^{2}+4x+? \right)+3{y}^{2}-6y=24+?$

To complete the square ${x}^{2}+4x+4={\left( x+2 \right)}^{2}$ add $4$ to the expression
$3\left( {x}^{2}+4x+4 \right)+3{y}^{2}-6y=24+?$

Since $3 \times 4$ was added to the left-hand side, also add $3 \times 4$ to the right-hand side
$3\left( {x}^{2}+4x+4 \right)+3{y}^{2}-6y=24+3 \times 4$

Use ${a}^{2}+2ab+{b}^{2}={\left( a+b \right)}^{2}$ to factor the expression
$3{\left( x+2 \right)}^{2}+3{y}^{2}-6y=24+3 \times 4$

Simplify the expression
$3{\left( x+2 \right)}^{2}+3{y}^{2}-6y=36$

Factor out $3$ from the expression
$3{\left( x+2 \right)}^{2}+3\left( {y}^{2}-2y \right)=36$

To complete the square, the same value needs to be added to both sides
$3{\left( x+2 \right)}^{2}+3\left( {y}^{2}-2y+? \right)=36+?$

To complete the square ${y}^{2}-2y+1={\left( y-1 \right)}^{2}$ add $1$ to the expression
$3{\left( x+2 \right)}^{2}+3\left( {y}^{2}-2y+1 \right)=36+?$

Since $3 \times 1$ was added to the left-hand side, also add $3 \times 1$ to the right-hand side
$3{\left( x+2 \right)}^{2}+3\left( {y}^{2}-2y+1 \right)=36+3 \times 1$

Use ${a}^{2}-2ab+{b}^{2}={\left( a-b \right)}^{2}$ to factor the expression
$3{\left( x+2 \right)}^{2}+3{\left( y-1 \right)}^{2}=36+3 \times 1$

Simplify the expression
$3{\left( x+2 \right)}^{2}+3{\left( y-1 \right)}^{2}=39$

Divide both sides of the equation by $3$
${\left( x+2 \right)}^{2}+{\left( y-1 \right)}^{2}=13$

The equation can be written in the form ${\left( x-p \right)}^{2}+{\left( y-q \right)}^{2}={r}^{2}$ , so it represents a circle with the radius $r=\sqrt{ 13 }$ and the center $\left( -2, 1\right)$
$\textnormal{Circle with the radius }r=\sqrt{ 13 }\textnormal{ and the center }\left( -2, 1\right)$

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