Pré-Vestibular ⇒ (Unesp) Geometria Analítica Tópico resolvido

[brunocbbc]

(Unesp) Se M=(5/2,0) é o ponto médio do
segmento cujos extremos são as interseções da
circunferência x²+y²+mx-y-4=0
com o eixo x, determine o centro dessa
circunferência.

Resposta

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(5/2, 1/2)

[petras]

brunocbbc,

[tex3]\mathsf{
A(x_A,0)\\
B(x_B,0)\\
\frac{x_A+x_B}{2}=\frac{5}{2}\implies x_A+x_B = 5\\
y = 0\implies x^2+0+mx-0-4=0 \therefore x^2+mx-4=0\\
x_A :x_B;raízes\implies m = -5 (x^2-Sx+P=0)\\
\therefore x^2+y^2-5x-y-4 = 0 \implies x^2-5x+y^2-y=4\\
x^2-5x+\frac{25}{4}+y^2-y+\frac{1}{4 }=4+\frac{25}{4}+\frac{1}{4}\implies (x-\frac{5}{2})^2+(y-\frac{1}{2})^2=13\\
\therefore \boxed{(\frac{5}{2}, \frac{1}{2})}\color{green}\checkmark












}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
A(x_A,0)\\
B(x_B,0)\\
\frac{x_A+x_B}{2}=\frac{5}{2}\implies x_A+x_B = 5\\
y = 0\implies x^2+0+mx-0-4=0 \therefore x^2+mx-4=0\\
x_A :x_B;raízes\implies m = -5 (x^2-Sx+P=0)\\
\therefore x^2+y^2-5x-y-4 = 0 \implies x^2-5x+y^2-y=4\\
x^2-5x+\frac{25}{4}+y^2-y+\frac{1}{4 }=4+\frac{25}{4}+\frac{1}{4}\implies (x-\frac{5}{2})^2+(y-\frac{1}{2})^2=13\\
\therefore \boxed{(\frac{5}{2}, \frac{1}{2})}\color{green}\checkmark












}[/tex3]