Respuesta :
Answer:
[tex]x^2+y^2-5y+x=0[/tex]
Step-by-step explanation:
Let [tex]a(x^2+y^2)+bx+cy+d=0[/tex] be the equation of the circle. We know that [tex]a[/tex] must be iqual [tex]1[/tex], due to the canonical equation of the circle is [tex](x-h)^{2}+(y-k)^{2}=r^{2}.[/tex], but we can think [tex]a[/tex] as an unknown for the moment.
In order to determine the equation of the circle we have to find the coefficients [tex]a,b,c,d[/tex]. In addition, we have that [tex](2,3), (-3,2), (0,0)[/tex] are point on this circle. Furthermore, If [tex]P=(x,y)[/tex] is any point in circle, then the following equations are satisfied:
[tex]\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}[/tex]
Here the unknowns are [tex]a,b,c,d[/tex]. 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
[tex]\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[/tex]
Calculating this determinant we obtain the equation:
[tex]13x^{2}+13y^2-65y+13x=0[/tex]
Simplifying:
[tex]x^2+y^2-5y+x=0[/tex]
