Chapter SH, Problem 9.8EP

To determine

To find: the equation of the circle using two end point of a Diameter.

Answer to Problem 9.8EP

The equation of the circle will be ${\left(\text{x+}\frac{3}{2}\right)}^{\text{2}}\text{+}{\left(\text{y-4}\right)}^{\text{2}}\text{=}\frac{29}{4}$ .

Explanation of Solution

Given:

Two end point of a Diameter are:

  $\left(1,3\right),\left(-4,5\right)$ .

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 $\left(1,3\right),\left(-4,5\right)$ .

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 $\text{radius}\text{.}$

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 $\left(1,3\right),\left(-4,5\right)$ .

  $\text{centre=}\frac{\text{1}}{\text{2}}\left(\text{1+}\left(-4\right)\right)\text{,}\frac{\text{1}}{\text{2}}\left(\text{3+5}\right)$ .

  $\Rightarrow \frac{\text{1}}{\text{2}}\left(-3\right)\text{,4}$ .

  $\Rightarrow \frac{-3}{2}\text{,4}$ .

Center= $-\frac{3}{2},4$ .

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.

  $\left(1,3\right)$ and $-\frac{3}{2},4$ .

  $\text{r=}\sqrt{{\left(\text{-}\frac{3}{2}\text{-1}\right)}^{\text{2}}\text{+}{\left(\text{4-3}\right)}^{\text{2}}}$ .

  $\Rightarrow \text{r=}\sqrt{{\left(\frac{\text{-5}}{\text{2}}\right)}^{\text{2}}\text{+1}}$ .

  $\Rightarrow \text{r=}\sqrt{\frac{25}{4}\text{+1}}$ .

  $\Rightarrow \text{r=}\sqrt{\frac{\text{29}}{\text{4}}}$ .

  $\text{r=}\frac{\sqrt{29}}{2}$ .

So, the equation of the circle will be:

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

  ${\left(\text{x-}\left(\text{-}\frac{\text{3}}{\text{2}}\right)\right)}^{\text{2}}\text{+}{\left(\text{y-4}\right)}^{\text{2}}\text{=}{\left(\frac{\sqrt{\text{29}}}{\text{2}}\right)}^{\text{2}}$ .

  ${\left(\text{x+}\frac{3}{2}\right)}^{\text{2}}\text{+}{\left(\text{y-4}\right)}^{\text{2}}\text{=}\frac{29}{4}$ .

Hence the equation of the circle will be ${\left(\text{x+}\frac{3}{2}\right)}^{\text{2}}\text{+}{\left(\text{y-4}\right)}^{\text{2}}\text{=}\frac{29}{4}$ .