Chapter SH, Problem 9.9EP

To determine

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

Answer to Problem 9.9EP

The equation of the circle will be ${\left(\text{x-2}\right)}^{\text{2}}\text{+}{\left(\text{y-}\frac{\text{5}}{\text{2}}\right)}^{\text{2}}\text{=}\frac{73}{4}$ .

Explanation of Solution

Given:

Two end points of the diameter of the circle:

  $\left(-2,4\right),\left(6,1\right)$ .

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 $\left(-2,4\right),\left(6,1\right)$ .

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 $\left(-2,4\right),\left(6,1\right)$ .

  $\text{centre=}\frac{\text{1}}{\text{2}}\left(\text{-2+}6\right)\text{,}\frac{\text{1}}{\text{2}}\left(\text{4+1}\right)$ .

  $\Rightarrow 2\text{,}\frac{5}{2}$ .

Center= $2\text{,}\frac{5}{2}$ .

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.

  $\left(-2,4\right)$ and $2\text{,}\frac{5}{2}$ .

  $\text{r=}\sqrt{{\left(2\text{-}\left(-2\right)\right)}^{\text{2}}\text{+}{\left(\frac{5}{2}\text{-4}\right)}^{\text{2}}}$ .

  $\Rightarrow \text{r=}\sqrt{\text{16+}\frac{9}{4}}$ .

  $\Rightarrow \text{r=}\sqrt{\frac{64+9}{4}}$ .

  $\Rightarrow \text{r=}\sqrt{\frac{73}{\text{4}}}$ .

  $\text{r=}\frac{\sqrt{73}}{2}$ .

So, the equation of the circle will be:

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

  ${\left(\text{x-2}\right)}^{\text{2}}\text{+}{\left(\text{y-}\frac{\text{5}}{\text{2}}\right)}^{\text{2}}\text{=}{\left(\frac{\sqrt{\text{73}}}{\text{2}}\right)}^{\text{2}}$ .

  ${\left(\text{x-2}\right)}^{\text{2}}\text{+}{\left(\text{y-}\frac{\text{5}}{\text{2}}\right)}^{\text{2}}\text{=}\frac{73}{4}$ .

Hence the equation of the circle will be ${\left(\text{x-2}\right)}^{\text{2}}\text{+}{\left(\text{y-}\frac{\text{5}}{\text{2}}\right)}^{\text{2}}\text{=}\frac{73}{4}$ .