Chapter SH, Problem 9.10EP

To determine

To find: the equation of the two Dimension sketch of gear using two end point of a Diameter.

Answer to Problem 9.10EP

The equation of the circle will be ${\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=29}$ .

Explanation of Solution

Given:

Two end points of the diameter of the gear is $\left(\text{-2,5}\right)\text{,}\left(\text{2,-5}\right)$ .

Concept used:

The standard form of the equation of a circle:

  ${\left(\text{x-h}\right)}^{\text{2}}\text{+}{\left(\text{y-k}\right)}^{\text{2}}{\text{=r}}^{\text{2}}$ .

Midpoint formula:

  $\frac{\text{1}}{\text{2}}\left({\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}\right)\text{,}\frac{\text{1}}{\text{2}}\left({\text{y}}_{\text{1}}{\text{+y}}_{\text{2}}\right)$ .

To calculate the $\text{radius}$ use the distance formula:

  $\text{r=}\sqrt{{\left({\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}\right)}^{\text{2}}\text{+}{\left({\text{y}}_{\text{2}}{\text{-y}}_{\text{1}}\right)}^{\text{2}}}$ .

Calculation:

The coordinate of the two end points of the diameter of the circle $\left(\text{-2,5}\right)\text{,}\left(\text{2,-5}\right)$ .

The standard form of the equation of a circle:

  ${\left(\text{x-h}\right)}^{\text{2}}\text{+}{\left(\text{y-k}\right)}^{\text{2}}{\text{=r}}^{\text{2}}$ .

Where $\text{h,k}$ are the center point and $\text{r}$ is $\text{radius}$ .

To establish the equation of circle the center point and the radius of the circle is required.

Since, endpoints of the diameter is given then the center will be mid- point and the radius will be the distance between endpoint and the center.

Midpoint formula:

  $\frac{\text{1}}{\text{2}}\left({\text{x}}_{\text{1}}{\text{+x}}_{\text{2}}\right)\text{,}\frac{\text{1}}{\text{2}}\left({\text{y}}_{\text{1}}{\text{+y}}_{\text{2}}\right)$ .

The two points here is $\left(\text{-2,5}\right)\text{,}\left(\text{2,-5}\right)$ .

  $\text{centre=}\frac{\text{1}}{\text{2}}\left(\text{-2+2}\right)\text{,}\frac{\text{1}}{\text{2}}\left(\text{5+}\left(\text{-5}\right)\right)$ .

  $\Rightarrow 0,0$ .

Center= $0,0$ .

To calculate the $\text{radius}$ use the distance formula:

  $\text{r=}\sqrt{{\left({\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}\right)}^{\text{2}}\text{+}{\left({\text{y}}_{\text{2}}{\text{-y}}_{\text{1}}\right)}^{\text{2}}}$ .

Let the two point be one end of the diameter and center of the circle.

  $\left(\text{-2,5}\right)$ and $0,0$ .

  $\text{r=}\sqrt{{\left(\text{0-}\left(\text{-2}\right)\right)}^{\text{2}}\text{+}{\left(\text{0-5}\right)}^{\text{2}}}$ .

  $\Rightarrow \text{r=}\sqrt{\text{4+25}}$ .

  $\text{r=}\sqrt{29}$ .

So, the equation of the circle will be:

  ${\left(\text{x-h}\right)}^{\text{2}}\text{+}{\left(\text{y-k}\right)}^{\text{2}}{\text{=r}}^{\text{2}}$

  ${\left(\text{x-0}\right)}^{\text{2}}\text{+}{\left(\text{y-0}\right)}^{\text{2}}\text{=}{\left(\sqrt{\text{29}}\right)}^{\text{2}}$ .

  ${\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=29}$ .

Hence the equation of the circle will be ${\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=29}$ .