Chapter 12.5, Problem 55E

To determine

To find: the standard form of the equation of the circle that passed through the given points.

Answer to Problem 55E

The general equation for the circle is:

  $x^2+y-4y-45=0.$

Explanation of Solution

Given:

Three points are $\left(0,9\right),\left(-7,2\right)\text{ and }\left(0,-5\right)$ .

Concept used:

The circle equation:

  $x^2+y+Dx+Ey+F=0.$

Where $\left(x,y\right)$ are the coordinates.

Calculation:

Let the circle equation be:

  $x^2+y+Dx+Ey+F=0.$

Three points are $\left(0,9\right),\left(-7,2\right)\text{ and }\left(0,-5\right)$ .

By putting each value of coordinate $x\text{ and }y$ in equation of circle:

  $\begin{array}{l}0+81+0+9E+F=0.\\ 49+4-7D+2E+F=0.\\ 0+25+0-5E+F=0.\end{array}$

Solving the equation by adding or eliminating the variable:

  $D=0,E=-4\text{ and }F=-45$ .

Hence, the general equation for the circle is:

  $x^2+y-4y-45=0.$