Question

How do find the standard equation of the circle with endpoints of a diameter $\left(3,6\right)$and $\left(-1,4\right)$?

Answer 1

Step-1: To find center of circle:

Standard equation of circle is given as-

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where $\left(h,k\right)$ is center points of circle and $r$ is radius.

Center point of circle can be determined by finding mid point of line segment joining $\left(3,6\right)$ and $\left(-1,4\right)$.

The mid point for line segment joining points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Calculate the mid point of line segment joining $\left(3,6\right)$ and $\left(-1,4\right)$.

$\begin{array}{rcl}M& =& \left(\frac{3+\left(-1\right)}{2},\frac{6+4}{2}\right)\\ & \Rightarrow & M=\left(\frac{2}{2},\frac{10}{2}\right)\\ \therefore M& =& \left(1,5\right)\end{array}$

Step-2: To find radius of circle:

Radius of circle can be determined by finding distance between the points $\left(3,6\right)$ and $\left(1,5\right)$

Distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

The distance between points $\left(3,6\right)$ and $\left(1,5\right)$ given as-

$\begin{array}{rcl}d& =& \sqrt{{\left(3-1\right)}^2+{\left(6-5\right)}^2}\\ & \Rightarrow & d=\sqrt{4+1}\\ \therefore d& =& \sqrt{5}\end{array}$

Step-3: To find standard equation of circle:

Substitute center points and radius of circle in equation $\left(1\right)$-

$$\begin{gathered} {\left(x-1\right)}^2+{\left(y-5\right)}^2={\left(\sqrt{5}\right)}^2 \\ {\left(x-1\right)}^2+{\left(y-5\right)}^2==5 \end{gathered}$$

Hence, the standard equation of circle is ${\left(x-1\right)}^{\mathbf{2}}\mathbf{+}{\left(y-5\right)}^{\mathbf{2}}\mathbf{=}\mathbf{=}\mathbf{5}$.