Chapter 1.4, Problem 25AYU

To determine

To find:

1. The centre $\left(h,k\right)$ and radius of the circle $x²+y²-2x-4y-4=0$

2. The graph of the circle

3. Intercepts of the circle, if any

Answer to Problem 25AYU

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

$x\text{-intercepts }=\sqrt{5}+1\text{ and}-\sqrt{5}+1;y\text{-intercepts }=2+\sqrt{8}\text{ and }2-\sqrt{8}$

Explanation of Solution

Given:

Equation of the circle $x²+y²-2x-4y-4=0$

Calculation:-

$x²+y²-2x-4y-4=0$

$\left(x²-2x\right)+\left(y²-4y\right)-4=0$

$\left(x²-2x\right)+\left(y²-4y\right)-4+4=0+4$   Add 4 to both sides

$\left(\text{x² <mo>−</mo> 2x <mo>+</mo> 1}\right)\text{ <mo>+</mo> }(\text{y² <mo>−</mo> 4y <mo>+</mo> 4}\text{y² <mo>−</mo> 4y <mo>+</mo> 4})\text{ <mo>=</mo> 4 <mo>+</mo> 5}$   Complete the square of each expression in parenthesis

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

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 $h=1,k=2$ and $r=3$.

The circle has centre $\left(1,2\right)$ and a radius 3 units. To graph the circle, first plot the centre $\left(1,2\right)$. Since the radius is 3 units, locate four points on the circle by plotting 3 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 $x$.

$\left(x-1\right)²+\left(\text{0 }-2\right)²=\left(3\right)²$

$\left(x-1\right)²+4=9$

$\left(x-1\right)²+4-4=9-4$   Subtract 4 from both sides

$\left(x-1\right)²=5$

$x-1=\pm \sqrt{5}$

$\text{If},x-1=\sqrt{5}$

$x-1+1=\sqrt{5}+1$   Add 1 to both sides

$x=\sqrt{5}+1$

$\text{If},x-1=-\sqrt{5}$

$x-1+1=- \sqrt{5}+1$   Add 1 to both sides

$x=-\sqrt{5}+1$

$x\text{-intercepts}$ are $\sqrt{5}+1$ and $-\sqrt{5}+1.$

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

$\left(0-1\right)²+\left(y-2\right)²=9$

$1+\left(y-2\right)²=9$

$\left(y-2\right)²+1-1=9-1$   Subtract 1 from both sides

$\left(y-2\right)²=8$

$y-2=\pm \sqrt{8}$

$\text{If }y-2=\sqrt{8}$

$y-2+2=\sqrt{8}+2$

$y=2+\sqrt{8}$

$\text{If }y-2=\text{ −}\sqrt{8}$

$y-2+2=\text{ −}\sqrt{8}+2$

$y=- \sqrt{8}+2$

$y\text{-intercepts}$ are $2+\sqrt{8}$ and $2-\sqrt{8}$.