Question

A equação da reta que passa pelos centros das
circunferências x² + y² - 8x + 2y +10 = 0 e
x² + y² + 4x -16y + 60 = 0 é
Valendo 30pts

Answer 1

Bom dia Cintia!

Solução!

Vamos pimeiramente determinar o centro das circunferencias.

Vou chamar centro1  e centro 2 indicadas por C1 e C2.

$x^{2} +y ^{2}-8x+2y+10=0\\\\\\ -2a=-8\\\\\ a= \dfrac{-8}{-2}\\\\\ a=4\\\\\\ -2b=2\\\\\ b= \dfrac{2}{-2} \\\\\ b=-1\\\\\\\ C1(4,-1)$

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$x^{2} +y^{2}+4x-16y+60=0\\\\\\\ -2a=4\\\\\ a= \dfrac{4}{-2}\\\\\ a=-2\\\\\\\\\\ -2b=-16\\\\\\ b= \dfrac{-16}{-2}\\\\\\ b=8\\\\\\\ C2(-2,8)$

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Vamos agora determinar o coeficiente angular da reta que passa pelo centro das circunferencias

$C1(4,-1)\\\\\\\ C2(-2,8)\\\\\\ m= \dfrac{C2y-C1y}{C2x-C1x}\\\\\\ m= \dfrac{8+1}{-2-4}\\\\\\ m= \dfrac{9}{-6}\\\\\\ m= -\dfrac{3}{2}$

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Equação da reta!

$C1(4,-1)\\\\\\\ m= -\dfrac{3}{2} \\\\\\\ y-yC1=m(x-xC1)\\\\\\\\\ y+1= -\frac{3}{2}(x-4)\\\\\\ y+1= -\frac{3x}{2}+ \frac{12}{2} \\\\\\\\ y+1= -\frac{3x+12}{2} \\\\\\\\ 2y+2=-3x+12\\\\\\ 3x+2y+2-12=0$

$\boxed{3x+2y-10=0~~\Rightarrow~~Eq:Forma~~geral}$

$\boxed{y= -\frac{3x}{2}-5~~ \Rightarrow~~Eq:Forma~~reduzida}$

Veja qual das duas formas estão no gabarito!

Bom dia!

Bons estudos!