Question
Find the center coordinates and radius of a circle for an equation written as: 3x2 + 3y2 - 6y = —12× + 24
69
likes347 views
Answer to a math question Find the center coordinates and radius of a circle for an equation written as: 3x2 + 3y2 - 6y = —12× + 24
Ali
4.490 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 $\left( -2, 1\right)$
$\textnormal{Circle with the radius }r=\sqrt{ 13 }\textnormal{ and the center }\left( -2, 1\right)$
Frequently asked questions (FAQs)
What is the domain of the function y = cos(2x) - tan(x) in terms of radians?
+
What is the value of arccot(3)
+
What is the maximum possible value of f(x) = x^2 - 4x + 3 over the interval [0, 5]?
+
New questions in Mathematics