Chapter 1.4, Problem 24AYU

To determine

To find:

1. The centre $\left(h,k\right)$ and radius of the circle $3\left(x+1\right)²+3\left(y-1\right)²=6$

2. The graph of the circle

3. Intercepts of the circle, if any

Answer to Problem 24AYU

$\text{Centre }\left(1,2\right);\text{ radius }= \sqrt{2};$

$x\text{-intercepts }=-2,0;y\text{-intercept }=2,0$

Explanation of Solution

Given:

Equation of the circle $3\left(x+1\right)²+3\left(y-1\right)²=6$

Calculation:-

$3\left(x+1\right)²+3\left(y-1\right)²=6$

$\left(x+1\right)²+\left(y-1\right)²=2$   Divide both sides by 3

Let us write the equation of the circle in standard form $\left(x+1\right)²+(y-1)²=(\sqrt{2})²$.

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

The comparison yields the information about the circle. We see that $h=-1,k=1$ and $r=\sqrt{2}$.

The circle has centre $\left(\mathrm{-1},1\right)$ and a radius $\sqrt{2}$ units. To graph the circle, first plot the centre $\left(\mathrm{-1},1\right)$. Since the radius is $\sqrt{2}$ units, locate four points on the circle by plotting $\sqrt{2}$ 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 $y=0$ and solve for $x$.

$\left(x+1\right)²+\left(0-1\right)²=2$

$\left(x+1\right)²+1=2$

$\left(x+1\right)²+1-1=2-1$   Subtract 1 from both sides

$\left(x+1\right)²=1$

$x+1=\pm 1$

$\text{If},x+1=1$

$x+1-1=1-1$   Subtract 1 from both sides

$x=0$

$\text{If},x+1=\text{ −}1$

$x+1-1=\text{ −}1-1$   Subtract 1 from both sides

$x=\text{ −}2$

$x\text{-intercepts}$ are $-2$ and 0.

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

$\left(0+1\right)²+\left(y-1\right)²=2$

$1+\left(y-1\right)²=2$

$\left(y-1\right)²+1-1=2-1$   Subtract 1 from both sides

$\left(y-1\right)²=1$

$y-1=\pm 1$

$\text{If }y-1=1$

$y-1+1=1+1$

$y=2$

$\text{If }y-1=\text{ −}1$

$y-1+1=\text{ −}1+1$

$y=0$

$y\text{-intercepts}$ are 2 and 0.