Chapter 1.8, Problem 121E

  $\left(a\right)$

To determine

To calculate: The radius of the circles and distance between their centers.

  $\begin{array}{c}\left(i\right){\left(x-2\right)}^2+{\left(y-1\right)}^2=9\\ {\left(x-6\right)}^2+{\left(y-4\right)}^2=16\end{array}$

  $\begin{array}{c}\left(ii\right)x^2+{\left(y-2\right)}^2=4\\ {\left(x-5\right)}^2+{\left(y-14\right)}^2=9\end{array}$

  $\begin{array}{c}\left(iii\right){\left(x-3\right)}^2+{\left(y+1\right)}^2=1\\ {\left(x-2\right)}^2+{\left(y-2\right)}^2=25\end{array}$

And interpret whether the circles intersect or not.

Answer to Problem 121E

The radius of the circles and distance between their centers is provided below,

  $\begin{array}{cccc}\text{Equation of circles}& \text{Radius}& \begin{array}{l}\text{Distance between}\\ \text{their centers}\end{array}& \text{Intersect}\\ \begin{array}{c}\left(i\right){\left(x-2\right)}^2+{\left(y-1\right)}^2=9\\ {\left(x-6\right)}^2+{\left(y-4\right)}^2=16\end{array}& \begin{array}{l}3\\ 4\end{array}& 5& \text{Yes}\\ \begin{array}{c}\left(ii\right)x^2+{\left(y-2\right)}^2=4\\ {\left(x-5\right)}^2+{\left(y-14\right)}^2=9\end{array}& \begin{array}{l}2\\ 3\end{array}& 13& \text{No}\\ \begin{array}{c}\left(iii\right){\left(x-3\right)}^2+{\left(y+1\right)}^2=1\\ {\left(x-2\right)}^2+{\left(y-2\right)}^2=25\end{array}& \begin{array}{l}1\\ 5\end{array}& 3.2& \text{Yes}\end{array}$

Explanation of Solution

Given information:

The pair of equation of circles,

  $\begin{array}{c}\left(i\right){\left(x-2\right)}^2+{\left(y-1\right)}^2=9\\ {\left(x-6\right)}^2+{\left(y-4\right)}^2=16\end{array}$

  $\begin{array}{c}\left(ii\right)x^2+{\left(y-2\right)}^2=4\\ {\left(x-5\right)}^2+{\left(y-14\right)}^2=9\end{array}$

  $\begin{array}{c}\left(iii\right){\left(x-3\right)}^2+{\left(y+1\right)}^2=1\\ {\left(x-2\right)}^2+{\left(y-2\right)}^2=25\end{array}$

Formula used:

The standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

Consider the equation,

  $\begin{array}{c}\left(i\right){\left(x-2\right)}^2+{\left(y-1\right)}^2=9\\ {\left(x-6\right)}^2+{\left(y-4\right)}^2=16\end{array}$ .

Rewrite the equation ${\left(x-2\right)}^2+{\left(y-1\right)}^2=9$ as,

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

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-2\right)}^2+{\left(y-1\right)}^2=3^2$ .

Here, $h=2,k=1\text{ and }r=3$ .

Therefore, center of circle is $\left(2,1\right)$ and radius is $3$ .

Next, rewrite the equation ${\left(x-6\right)}^2+{\left(y-4\right)}^2=16$ as,

  ${\left(x-6\right)}^2+{\left(y-4\right)}^2=4^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-6\right)}^2+{\left(y-4\right)}^2=4^2$ .

Here, $h=6,k=4\text{ and }r=4$ .

Therefore, center of circle is $\left(6,4\right)$ and radius is $4$ .

Now, distance between the centers of the circle is computed below,

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $A\left(6,4\right)$ and $B\left(2,1\right)$ .

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(2-6\right)}^2+{\left(1-4\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

Now, sum of radius of two circles is $3+4=7$ .

When the distance between the two centers of the circle is less than sum of radius of two circles then the two circles intersect each other.

Therefore, the circles ${\left(x-2\right)}^2+{\left(y-1\right)}^2=9\text{ and }{\left(x-6\right)}^2+{\left(y-4\right)}^2=16$ intersect each other.

Consider the equation,

  $\begin{array}{c}\left(ii\right)x^2+{\left(y-2\right)}^2=4\\ {\left(x-5\right)}^2+{\left(y-14\right)}^2=9\end{array}$ .

Rewrite the equation $x^2+{\left(y-2\right)}^2=4$ as,

  ${\left(x-0\right)}^2+{\left(y-2\right)}^2=2^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-0\right)}^2+{\left(y-2\right)}^2=2^2$ .

Here, $h=0,k=2\text{ and }r=2$ .

Therefore, center of circle is $\left(0,2\right)$ and radius is $2$ .

Next, rewrite the equation ${\left(x-5\right)}^2+{\left(y-14\right)}^2=9$ as,

  ${\left(x-5\right)}^2+{\left(y-14\right)}^2=3^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-5\right)}^2+{\left(y-14\right)}^2=3^2$ .

Here, $h=5,k=14\text{ and }r=3$ .

Therefore, center of circle is $\left(5,14\right)$ and radius is $3$ .

Now, distance between the centers of the circle is computed below,

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $A\left(0,2\right)$ and $B\left(5,14\right)$ .

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(5-0\right)}^2+{\left(14-2\right)}^2}\\ =\sqrt{25+144}\\ =\sqrt{169}\\ =13\end{array}$

Now, sum of radius of two circles is $2+3=5$ .

When the distance between the two centers of the circle is less than sum of radius of two circles then the two circles intersect each other.

Therefore, the circles $x^2+{\left(y-2\right)}^2=4\text{ and }{\left(x-5\right)}^2+{\left(y-14\right)}^2=9$ do not intersect each other.

Consider the equation,

  $\begin{array}{c}\left(iii\right){\left(x-3\right)}^2+{\left(y+1\right)}^2=1\\ {\left(x-2\right)}^2+{\left(y-2\right)}^2=25\end{array}$ .

Rewrite the equation ${\left(x-3\right)}^2+{\left(y+1\right)}^2=1$ as,

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

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-3\right)}^2+{\left(y-\left(-1\right)\right)}^2=1^2$ .

Here, $h=3,k=-1\text{ and }r=1$ .

Therefore, center of circle is $\left(3,-1\right)$ and radius is $1$ .

Next, rewrite the equation ${\left(x-2\right)}^2+{\left(y-2\right)}^2=25$ as,

  ${\left(x-2\right)}^2+{\left(y-2\right)}^2=5^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-2\right)}^2+{\left(y-2\right)}^2=5^2$ .

Here, $h=2,k=2\text{ and }r=2$ .

Therefore, center of circle is $\left(2,2\right)$ and radius is $5$ .

Now, distance between the centers of the circle is computed below,

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $A\left(3,-1\right)$ and $B\left(2,2\right)$ .

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(2-3\right)}^2+{\left(2+1\right)}^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\\ =3.2\end{array}$

Now, sum of radius of two circles is $1+5=6$ .

When the distance between the two centers of the circle is less than sum of radius of two circles then the two circles intersect each other.

Therefore, the circles ${\left(x-3\right)}^2+{\left(y+1\right)}^2=1\text{ and }{\left(x-2\right)}^2+{\left(y-2\right)}^2=25$ intersect each other.

Therefore, the above results are summarized as,

  $\begin{array}{cccc}\text{Equation of circles}& \text{Radius}& \begin{array}{l}\text{Distance between}\\ \text{their centers}\end{array}& \text{Intersect}\\ \begin{array}{c}\left(i\right){\left(x-2\right)}^2+{\left(y-1\right)}^2=9\\ {\left(x-6\right)}^2+{\left(y-4\right)}^2=16\end{array}& \begin{array}{l}3\\ 4\end{array}& 5& \text{Yes}\\ \begin{array}{c}\left(ii\right)x^2+{\left(y-2\right)}^2=4\\ {\left(x-5\right)}^2+{\left(y-14\right)}^2=9\end{array}& \begin{array}{l}2\\ 3\end{array}& 13& \text{No}\\ \begin{array}{c}\left(iii\right){\left(x-3\right)}^2+{\left(y+1\right)}^2=1\\ {\left(x-2\right)}^2+{\left(y-2\right)}^2=25\end{array}& \begin{array}{l}1\\ 5\end{array}& 3.2& \text{Yes}\end{array}$

  $\left(b\right)$

To determine

To explain: Whether the circles intersect each other or not provided their radius and distance between their centers.

Answer to Problem 121E

When the distance between the two centers of the circle is less than sum of radius of two circles then the two circles intersect each other.

Explanation of Solution

Given information:

The pair of equation of circles.

Consider the equation,

  ${\left(x-2\right)}^2+{\left(y-1\right)}^2=9\text{ and }{\left(x-6\right)}^2+{\left(y-4\right)}^2=5$ . Plot the circles on the Cartesian plane. Red curve represent the first circle and the blue curve represent the second circle.

  

Rewrite the equation ${\left(x-2\right)}^2+{\left(y-1\right)}^2=9$ as,

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

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-2\right)}^2+{\left(y-1\right)}^2=3^2$ .

Here, $h=2,k=1\text{ and }r=3$ .

Therefore, center of circle is $\left(2,1\right)$ and radius is $3$ .

Next, rewrite the equation ${\left(x-6\right)}^2+{\left(y-4\right)}^2=16$ as,

  ${\left(x-6\right)}^2+{\left(y-4\right)}^2={\left(\sqrt{5}\right)}^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-6\right)}^2+{\left(y-4\right)}^2=4^2$ .

Here, $h=6,k=4\text{ and }r=\sqrt{5}$ .

Therefore, center of circle is $\left(6,4\right)$ and radius is $\sqrt{5}$ .

Now, distance between the centers of the circle is computed below,

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $A\left(6,4\right)$ and $B\left(2,1\right)$ .

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(2-6\right)}^2+{\left(1-4\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

Now, sum of radius of two circles is $3+\sqrt{5}=5.23$ .

When the distance between the two centers of the circle is less than sum of radius of two circles then the two circles intersect each other.

Therefore, the circles ${\left(x-2\right)}^2+{\left(y-1\right)}^2=9\text{ and }{\left(x-6\right)}^2+{\left(y-4\right)}^2=5$ intersect each other.

Now, in general terms if d is the distance between the center of two circles with radius $r_1$ and $r_2$ then two circles intersect each other at two points if distance between the two centers of the circle is less than sum of radius of two circles that is $d<r_1+r_2$ .

When distance between the two centers of the circle is equal to sum of radius of two circles that is $d=r_1+r_2$ , then two circles touch each other at one point.

When distance between the two centers of the circle is equal to sum of radius of two circles that is $d>r_1+r_2$ , then two circles do not intersect each other.