Chapter 1, Problem 6CT

To determine

To write: standard form of equation of circle given below:

  

Answer to Problem 6CT

  ${\left(x-1\right)}^2+{\left(y-3\right)}^2=16$

Explanation of Solution

Concept Used:

The standard form of equation of circle with $\left(h,k\right)$ and radius r is:

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

Distance Formula: Distance between the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Midpoint Formula: Midpoint of the line segment joining the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

  $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

From the given graph it can be observed that the centre of the circle lies on the mid-point of line joining the points $\left(-3,3\right)$ and $\left(5,3\right)$ .

In the above midpoint formula substitute $\left(x_1,y_1\right)=\left(-3,3\right)$ and $\left(x_2,y_2\right)=\left(5,3\right)$ .

  $\begin{array}{c}C=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{-3+5}{2},\frac{3+3}{2}\right)\\ =\left(\frac{2}{2},\frac{6}{2}\right)\\ =\left(1,3\right)\end{array}$

The radius of the given circle is the distance between the centre of circle and any one point on its circumference, say $\left(5,3\right)$ .

In the above distance formula substitute $\left(x_1,y_1\right)=\left(1,3\right)$ and $\left(x_2,y_2\right)=\left(5,3\right)$ .

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(5-1\right)}^2+{\left(3-3\right)}^2}\\ =\sqrt{4^2+0^2}\end{array}$

  $\begin{array}{c}=\sqrt{16}\\ =4\end{array}$

Thus, the equation of circle in standard form is written as:

  $\begin{array}{c}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-1\right)}^2+{\left(y-3\right)}^2=4^2\\ {\left(x-1\right)}^2+{\left(y-3\right)}^2=16\end{array}$