Chapter 9.9, Problem 12P

To determine

To determine the equation of a circle that passes through the points $(-6,-3)$ , $(1,4)$ , $(2,3)$

Answer to Problem 12P

  $x^2+y^2+4x-21=0$ is the answer.

Explanation of Solution

Given:

a circle that passes through the points $(-6,-3)$ , $(1,4)$ , $(2,3)$

Formula Used:

The equation of a circle is:

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

Calculation:

Given the circle that passes through the points $(-6,-3)$ , $(1,4)$ , $(2,3)$

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

Hence, for the point $(-6,-3)$ :

  ${(-6)}^2+{(-3)}^2-6C-3D+E=0$

  $36+9-6C-3D+E=0$

  $-6C-3D+E=-45$

For the point $(1,4)$ :

  ${(1)}^2+{(4)}^2+C+4D+E=0$

  $1+16+C+4D+E=0$

  $C+4D+E=-17$

For the point $(2,3)$ :

  ${(2)}^2+{(3)}^2+2C+3D+E=0$

  $4+9+2C+3D+E=0$

  $2C+3D+E=-13$

Subtracting the equation $C+4D+E=-17$ from $-6C-3D+E=-45$ ; we get:

  $-7C-7D-28$

  $C+D=4$

  $D=4-C$

Subtracting the equation $2C+3D+E=-13$ from $C+4D+E=-17$ ; we get:

  $-C+D=-4$

Substituting the value of D in the equation $-C+D=-4$ , we get:

  $-C+4-C=-4$

  $-2C=-8$

  $C=4$

Putting the value of D in the equation $D=4-C$ , we get:

  $D=0$

Putting the values of C and D in the equation $2C+3D+E=-13$ , we get:

  $2\times 4+3\times 0+E=-13$

  $8+E=-13$

  $E=-21$

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

  $x^2+y^2+4x+0-21=0$

  $x^2+y^2+4x-21=0$

Hence, the answer is $x^2+y^2+4x-21=0$

Conclusion:

  $x^2+y^2+4x-21=0$ is the solution.