Question

alguien que la desarrolle es la ecuacion de una circunferencia
(x-h)^2 + (y-k)^2=r^2
(x+2)^2 + (y+4)^2=(40/13)^2

Answer 1

$\left(x+2\right)^2+\left(y+4\right)^2=\left(\frac{40}{13}\right)^2$
$\mathrm{Aplicar\:la\:siguiente\:regla\:de\:productos\:notables}:\quad \left(a+b\right)^2=a^2+2ab+b^2$
$x^2+4x+y^2+8y+20=\left(\frac{40}{13}\right)^2$
$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$
$x^2+4x+y^2+8y+20=\frac{40^2}{13^2}$
$x^2+4x+y^2+8y+20=\frac{1600}{169}$
$\mathrm{Multiplicar\:a\:ambos\:lados\:por\:}169:$
$169\cdot (x^2+4x+y^2+8y+20)=\left\frac{1600}{169}\right\cdot (169)$
$169 x^{2} +169 y^{2} +676x+1352y+3380=1600$
$169 x^{2} +169 y^{2} +676x+1352y+1780=0$