Chapter 1.7, Problem 9E

(a)

To determine

To prove: every point inside the circle ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4$ is inside the circle $x^2+y^2=41$ .

Explanation of Solution

Given information: ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4$

Proof:

On the graph we can see that the claim is correct, let’s prove it formally.

Assume x and y are real numbers such that.

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-2\right)}^2\le 4\\ \Downarrow \\ {\left(x-3\right)}^2\le 4\\ {\left(y-2\right)}^2\le 4\\ \Downarrow \\ -2\le x-3\le 2\\ -2\le y-2\le 2\\ \Downarrow \\ 1\le x\le 5\\ 0\le y\le 4\\ \Downarrow \\ x^2+y^2=25+16=41\end{array}$

The proof is done.

  

(b)

To determine

To prove: if $\left(x,y\right)$ is inside the circle ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4,$ then $x-6<3y$ .

Explanation of Solution

Given information: ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4,$

Proof:

On the graph we can see that the claim is correct, let’s prove it formally.

Assume x and y are real numbers such that.

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-2\right)}^2\le 4\\ \Downarrow \\ {\left(x-3\right)}^2\le 4\\ {\left(y-2\right)}^2\le 4\\ \Downarrow \\ -2\le x-3\le 2\\ -2\le y-2\le 2\\ \Downarrow \\ 1\le x\le 5\\ 0\le y\le 4\\ \Downarrow \\ -5\le x-6\le -1\\ 0\le 3y\le 12\\ \Downarrow \\ x-6\le -1<0\le 3y\end{array}$

The proof is done.

  

(c)

To determine

To prove: every point inside the circle ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4$ is inside the circle ${\left(x-5\right)}^2+{\left(y+1\right)}^2=25$ .

Explanation of Solution

Given information: ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4$ .

Proof:

On the graph we can see that the claim is correct, let’s find a counterexample.

Let $\left(x,y\right)\text{ be }\left(2,3,5\right),$ then

  ${\left(2-3\right)}^2+{\left(3\cdot 5-2\right)}^2=1+1\cdot 5^2=3\cdot 25\le 4$ .

It is in the first circle.

  ${\left(2-5\right)}^2+{\left(3\cdot 5+1\right)}^2=9+20\cdot 25=29\cdot 25>25$

It is not in the second circle.

  

(d)

To determine

To prove: if $\left(x,y\right)$ is inside the circle ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4$ is or the circle ${\left(x+1\right)}^2+{\left(y-3\right)}^2=9,$ then x + y > 3.

Explanation of Solution

Given information: ${\left(x-3\right)}^2+{\left(y-2\right)}^2=4$ .

Proof:

On the graph we can see that the claim is correct, let’s find a counterexample.

Let $\left(x,y\right)\text{ be }\left(0,4\right),$ then

  ${\left(0+1\right)}^2+{\left(4-3\right)}^2=1+1=2\le 9$ .

It is in the second circle.

But

  $0+4=4>3$

  

(e)

To determine

To prove: if $\left(x,y\right)$ is inside the circle ${\left(x-5\right)}^2+{\left(y-1\right)}^2=17$ , then $y<2$ or $xy>0$ .

Explanation of Solution

Given information: ${\left(x-5\right)}^2+{\left(y-1\right)}^2=17$ .

Proof:

On the graph we can see that the claim is correct, let’s prove it formally.

Assume x and y are real numbers such that.

  $\begin{array}{l}{\left(x-5\right)}^2+{\left(y-1\right)}^2\le 17\\ \Downarrow \\ {\left(x-5\right)}^2\le \sqrt{17}\\ {\left(y-1\right)}^2\le \sqrt{17}\\ \Downarrow \\ -\sqrt{17}\le x-5\le \sqrt{17}\\ -\sqrt{17}\le y-1\le \sqrt{17}\\ \Downarrow \\ -4.2\le x-5\le 4.1\\ -4.2\le y-1\le 4.1\\ \Downarrow \\ 0.8\le x\le 9.1\\ -3.2\le y\le 5.1\\ \Downarrow \\ -5\le x-6\le -1\\ 0\le 3y\le 12\\ \Downarrow \\ \left(y<2\vee y>2\right)\wedge x>0\\ \Downarrow \\ y<2\vee xy>0\end{array}$

The proof is done.

  

(f)

To determine

To prove: there is a unique real number x such that for all real numbers y , if y < x then x + y is negative and x − y is positive.

Explanation of Solution

Given information: there is a unique real number x such that for all real numbers y, y < x.

Proof:

On the graph we can see that the claim is correct, let’s find a counterexample.

Let $\left(x_1,y_1\right)\text{ be }\left(-2,-4\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(2,-4\right),$ then

  $\begin{array}{l}-4<-2\text{ and }-4<2\\ -2+\left(-4\right)=-6<0\text{ and }2+\left(-4\right)=-2<0\\ -2-\left(-4\right)=2>0\text{ and }2-\left(-4\right)=6>0\end{array}$

  $\left(x,y\right)$ exists but it is not unique.