Question

Linear Algebra Question

Answer 1

Answer:

$x^2+y^2-5y+x=0$

Step-by-step explanation:

Let $a(x^2+y^2)+bx+cy+d=0$ be the equation of the circle. We know that $a$ must be iqual $1$, due to the canonical equation of the circle is $(x-h)^{2}+(y-k)^{2}=r^{2}.$, but we can think $a$ as an unknown for the moment.

In order to determine the equation of the circle we have to find the coefficients $a,b,c,d$. In addition, we have that $(2,3), (-3,2), (0,0)$ are point on this circle. Furthermore, If $P=(x,y)$ is any point in circle, then the following equations are satisfied:

$\begin{cases}a(x^2+y^2)+bx+cy+d=0\\a(2^{2}+3^{2})+b(2)+c(3)+d=0\\a((-3)^{2}+2^{2})+b(-3)+c(2)+d=0\\a(0^{2}+0^{0})+b(0)+c(0)+d=0\end{cases}$

Here the unknowns are $a,b,c,d$.  According with a result of linear algebra, the system has a nontrivial solution if and only if the determinant of the matrix of coefficients is zero, then we have that

$\left \lvert \begin{array}{cccc}x^{2}+y^{2}&x &y&1\\(2^{2}+3^{2})&2&3&1\\((-3)^{2}+2^{2})&-3&2&1\\(0^{2}+0^{2})&0&0&1 \end{array} \right \rvert =0$

Calculating this determinant we obtain the equation:

$13x^{2}+13y^2-65y+13x=0$

Simplifying:

$x^2+y^2-5y+x=0$