Answer to Question #236803 in Algebra for moe

Question

You are given the circle $x^{2}+y^{2}+ax+by+c=0$, where a, b and c are some constants. Given that this circle passes through the points (4,2); (0,3) and (3,-2).

Which of the following are true?

1. The system of equations connecting a, b and c is $\begin{array}{cccc} 4a& +2b& +c&= -20\\ & +3b & +c& =-9 \\ 3a& -2b&+c&=-13 \end{array}$
2. The determinant of the coefficient matrix for the system connecting a, b and c is 17
3. With Cramer's rule, the value of a is $a=\frac{\left| \begin{array}{cccc}-20 & 2& 1\\ -9 & 3& 1\\ -13 & -2& 1 \end{array} \right |}{17}.$
4. With Cramer's rule, the value of b is $b=\frac{\left| \begin{array}{cccc}-20 & 2& 1\\ -9 & 3& 1\\ -13 & -2& 1 \end{array} \right |}{17}.$

Answer 1

firstly putting the values of points one by one in the given equation points are  (4,2), (0,3) and (3,-2).

$x^{2}+y^{2}+ax +by +c=0$

putting value (4,2) in equation we get

$16+4+4a+2b+c=0\\4a+2b+c=-20$

putting value (0,3) in equation we get

$0+9+0+3b+c=0\\3b+c=-9$

putting value (3,-2) in equation we get

$9+4+3a-2b+c=0$

$3a-2b+c=-13$

this shows that first option is correct

2 .after finding the determinant of matrix which is 17 second option is also correct

3 . according to cramer's rule the value of a is correct so third option is also correct

4 . fourth option is violating cramer's rule because it has its cofficient of b in the determinant