Chapter 9, Problem 5CR

To determine

To find: The standard form of the equation of the circle with the given characteristics.

Answer to Problem 5CR

The standard form of the equation of the circle is ${(x-2)}^2+{(y-4)}^2=13$ .

Explanation of Solution

Given information: Center: $(2,4)$ ; diameter: $2\sqrt{13}$

Calculations:

The standard form of the equation of the circle with center $(h,k)$ and radius r is given by:

  ${(x-h)}^2+{(y-k)}^2=r^2$

Here, center at origin.

i.e., $(h,k)\equiv (2,4)$

So, equation of the circle becomes ${(x-2)}^2+{(y-4)}^2=r^2$

Now, finding for radius $r$ :

  $\begin{array}{l}radius=\frac{diameter}{2}\\ \Rightarrow r=\frac{2\sqrt{13}}{2}\\ \Rightarrow r=\sqrt{13}\end{array}$

So, equation of the circle:

  $\begin{array}{}$ {(x-2)}^2+{(y-4)}^2={(\sqrt{13})}^2$$ \Rightarrow {(x-2)}^2+{(y-4)}^2=13$\end{array}$

Hence, the standard form of the equation of the circle is ${(x-2)}^2+{(y-4)}^2=13$ .