Chapter 8.3, Problem 40E

Solution

Write the standard form of circle that passes through the given point and it’s center is the origin.

  $x^2+y^2=169$

Given information:

Given point is:

  $\left(5,-12\right)$ .

Formula used:

Standard form of equation of a circle with radius ‘ r’ and center $\left(h,k\right)$ is

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

Calculation:

In order to write the standard form of circle that passes through the given point and it’s center is the origin, use the standard form of the equation of the circle:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , with radius ‘ r’ and center $\left(h,k\right)$ .

Note that the centre of given circle is origin and is passing through point $\left(5,-12\right)$ , i.e. here

  $\left(h,k\right)=\left(0,0\right)$

Thus, the equation becomes:

  $\begin{array}{c}{\left(x-0\right)}^2+{\left(y-0\right)}^2=r^2\\ x^2+y^2=r^2\end{array}$

Now, to find ‘ r’ use the given point through which the circle passes through by putting

  $\left(x,y\right)=\left(5,-12\right)$ in above equation as:

  $\begin{array}{c}{x}^2+y^2=r^2\\ 5^2+{\left(-12\right)}^2=r^2\\ r^2=25+144\\ r^2=169\\ r=\sqrt{169}\\ \Rightarrow r=13\end{array}$

Putting the value of radius, ‘ r’ :

  $\begin{array}{l}{x}^2+y^2={\left(13\right)}^2\\ x^2+y^2=169\end{array}$

Thus, the standard form of the equation of circle with given information is:

  $x^2+y^2=169$