Chapter 9, Problem 4CR

To determine

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

Answer to Problem 4CR

The standard form of the equation of the circle is $x^2+y^2=289$ .

Explanation of Solution

Given information: Center at origin; point on the circle: $(8,-15)$

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., $(h,k)\equiv (0,0)$

So, equation of the circle becomes $x^2+y^2=r^2$

Now, finding for $r^2$ :

Since, point $(8,-15)$ lies on the circle.

So, putting $x=8$ and $y=-15$ in the equation of circle:

  $\begin{array}{l}$ {(8)}^2+{(-15)}^2=r^2$\Rightarrow r^2=64+225\\ \Rightarrow r^2=289\end{array}$

Hence, the standard form of the equation of the circle is $x^2+y^2=289$ .