Question

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

69

likes
347 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
4.4
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 relationship between the coefficient 'a' and the width and direction of the parabola in the function 𝑦 = ax^2?
+
What is the value of $\frac{5\pi}{3}$ angle in degrees?
+
What is the value of y when x = 10 in the logarithmic function y = log$base 2$$x + 3$?
+