Chapter 9.9, Problem 11P

To determine

To determine the equation of a circle that passes through the points $(2,-1)$ , $(4,-3)$ and $(0,-3)$

Answer to Problem 11P

  $x^2+y^2-4x+6y+12=0$ is the solution.

Explanation of Solution

Given:

a circle that passes through the points $(2,-1)$ , $(4,-3)$ and $(0,-3)$

Formula Used:

The equation of a circle is:

  $x^2+y^2+Cx+Dy+E=0$ where C,D and E are the co-efficient.

Calculation:

Given the circle that passes through the points $(2,-1)$ , $(4,-3)$ , $(0,-3)$

We need to calculate the co-efficient for each of the three points by substituting the values of x and y for each point.

Hence, for the point $(2,-1)$ :

  $\begin{array}{l}$ {(2)}^2+{(-1)}^2+2C-D+E=0$36+9-6C-3D+E=0\\ 2C-D+E=-5\end{array}$

For the point $(4,-3)$ :

  $\begin{array}{l}$ {(4)}^2+{(-3)}^2+4C-3D+E=0$25+4C-3D+E=0\\ 4C-3D+E=-25\end{array}$

For the point $(0,-3)$ :

  $\begin{array}{l}$ {(0)}^2+{(-3)}^2+0\times C-3D+E=0$9-3D+E=0\\ E=3D-9\end{array}$

Solving the equation $4C-3D+E=-25$ and $2C-D+E=-5$ we have:

  $-2C+2D=20$

  $-C+D=10$

  $D=10+C$

Putting the value of D in the equation $E=3D-9$ , we have:

  $E=3(10+C)-9$

  $E=3C+21$

Now, putting the values of D and E in terms of C in the equation $2C-D+E=-5$ ; we have:

  $\begin{array}{l}2C-(10+C)+(21+3C)=-5\\ 2C-10-C+21+3C=-5\\ 4C+11=-5\\ 4C=-16\\ C=-4\end{array}$

Substituting the value of C in the equation $D=10+C$ for getting the value of D, we get:

  $D=10-4$

  $D=6$

Substituting the value of C in the equation $E=3C+21$ for getting the value of E, we get:

  $\begin{array}{l}E=3\times -3+21\\ E=12\end{array}$

Now, putting the values of C, D and E in the general equation of the circle, we get:

  $x^2+y^2+(-4\times x)+6\times y+12=0$

  $x^2+y^2-4x+6y+12=0$

Hence, the answer is $x^2+y^2-4x+6y+12=0$

Conclusion:

  $x^2+y^2-4x+6y+12=0$ is the solution.