Chapter 1.4, Problem 31AYU

To determine

To find:

1. The centre $\left(h,k\right)$ and radius of the circle $2x²+2y²-12x+8y-24=0$

2. The graph of the circle

3. Intercepts of the circle, if any

Answer to Problem 31AYU

$\text{Centre }\left(3,-2\right);\text{ radius }=5;$

$x\text{-intercepts are }3+\sqrt{21}\text{ and }3-\sqrt{21}$

$y\text{-intercepts are 2 and }-6$

Explanation of Solution

Given:

Equation of the circle $2x²+2y²-12x+8y-24=0$

Calculation:-

$2x²+2y²-12x+8y-24=0$

$x²+y²-6x+4y-12=0$

$(x²-6x)+(y²+4y)-12=0$

$\left(x²-6x\right)+\left(y²+4y\right)-12+12=0+12$   Add 12 to both sides

$\left(x²-6x+9\right)+\left(y²+4y+4\right)=12+13$   Complete the square of each expression in parenthesis

$\left(x-3\right)²+\left(y+2\right)²=\left(5\right)²$

Compare this equation with the equation $\left(x-h\right)²+\left(y-k\right)²=r²$.

The comparison yields the information about the circle. We see that $h=3,k=-2$ and $r=5$.

The circle has centre $\left(3,-2\right)$ and a radius 5 units. To graph the circle, first plot the centre $\left(3,-2\right)$. Since the radius is 5 units, locate four points on the circle by plotting 5 units to the left, to the right, up and down from the centre. These four points can be used to sketch the graph.

To find the $x\text{-intercepts}$, if any, let $y\text{= }0$ and solve for $x$.

$\left(x-3\right)²+\left(0+2\right)²=\left(5\right)²$

$\left(x-3\right)²+4=25$

$\left(x-3\right)²+4-4=25-4$   Subtract 4 from both sides

$\left(x-3\right)²=21$

$x-3=\pm \sqrt{21}$

$x-3+3=\pm \sqrt{21}+3$

$x=3\pm \sqrt{21}$

$x\text{-intercepts}$ are $3+\sqrt{21}$ and $3-\sqrt{21}$.

To find $y\text{-intercepts}$, if any, let $x=0$ and solve for $y$.

$\left(0-3\right)²+\left(y+2\right)²=25$

$9+\left(y+2\right)²=25$

$\left(y+2\right)²+9-9=25-9$   Subtract 9 from both sides

$\left(y+2\right)²=16$

$y+2=\pm 4$

$\text{If }y+2=4$

$y+2-2=4-2$

$y=2$

$\text{If }y+2=-4$

$y+2-2=-4-2$   Subtract 2 from both sides

$y=-6$

$y\text{-intercepts}$ are 2 and $-6$.