Respuesta :

WojtekR
You are right. In circle equation we have =r² so the correct answer will be C od D.  But we have to calculate the center of a circle.

[tex]x^2+y^2-8x-6y+24=0\quad|-24\\\\(x^2-8x)+(y^2-6y)=-24[/tex]

Complete the square:

[tex]x^2-8x+a^2\implies a=\dfrac{8x}{2x}=4\implies x^2-8x+4^2\\\\\\y^2-6y+b^2\implies b=\dfrac{6y}{2y}=3\implies y^2-6y+3^2[/tex]

And we have:

[tex](x^2-8x)+(y^2-6y)=-24\\\\(x^2-8x+4^2)-4^2+(y^2-6y+3^2)-3^2=-24\\\\ (x-4)^2+(y-3)^2-16-9=-24\\\\(x-4)^2+(y-3)^2-25=-24\quad|+25\\\\\boxed{(x-4)^2+(y-3)^2=1}[/tex]

So the answer is C (the same left side of equation).

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