Chapter 12, Problem 29CR

a.

To determine

To find: The distance between centers of two circles.

Answer to Problem 29CR

The distance $OP$ is $3$ .

Explanation of Solution

Given information:

The radius of circle with center Ois1y1y3$0.7$ .

The radius of circle with center Pis $1.1$ .

  $\overleftrightarrow{AB}$ is common internal tangent.

Side $AB=2.4$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Usecommon tangent procedure.

ABXO is a rectangle.

If AB = 2.4 then OX = 2.4

If AO = 0.7 then BX = 0.7

  $\begin{array}{l}PX=PB+BX\\ PX=1.1+0.7=1.8\end{array}$

Side OP can be calculated by applying Pythagoras Theorem.

In right angled triangle OXP, we get

  $\begin{array}{l}{\left(OP\right)}^2={\left(PX\right)}^2+{\left(OX\right)}^2\\ {\left(OP\right)}^2={\left(1.8\right)}^2+{\left(2.4\right)}^2\\ {\left(OP\right)}^2=3.24+5.76\\ {\left(OP\right)}^2=9\\ OP=\sqrt{9}=3\end{array}$

b.

To determine

To calculate: The distance between two circles.

Answer to Problem 29CR

The distanceis $1.2$ .

Explanation of Solution

Given information:

The radius of circle with center Ois$0.7$ .

The radius of circle with center P is $1.1$ .

  $\overleftrightarrow{AB}$ is common internal tangent.

Side $AB=2.4$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Usecommon tangent procedure.

ABXO is a rectangle.

If AB = 2.4 then OX = 2.4

If AO = 0.7 then BX = 0.7

  $\begin{array}{l}PX=PB+BX\\ PX=1.1+0.7=1.8\end{array}$

Side OP can be calculated by applying Pythagoras Theorem.

In right angled triangle OXP, we get

  $\begin{array}{l}{\left(OP\right)}^2={\left(PX\right)}^2+{\left(OX\right)}^2\\ {\left(OP\right)}^2={\left(1.8\right)}^2+{\left(2.4\right)}^2\\ {\left(OP\right)}^2=3.24+5.76\\ {\left(OP\right)}^2=9\\ OP=\sqrt{9}=3\end{array}$

Distance between two circles = OP − (Radius of circle with center O + Radius of circle with center P)

Distance between two circles $=3-\left(0.7+1.1\right)=3-1.8=1.2$