Question

Determine el centro y el radio de la circunferencia que satisface la ecuación dada
x^2+y^2=12
x^2+y^2-16=0
x^2+y^2-4x+6y=0
x^2+y^2-10x+2y+22=0

Answer 1

Explicación paso a paso:

$x ^{2} + y {}^{2} = 12$

$x ^{2} + y ^{2} + dx + ey + f = 0$

$c = ( \frac{ - d}{2} ; \frac{ - e}{2} )$

$c = ( \frac{ - 0}{2} ; \frac{ - 0}{2} )$

$c = 0;0$

$r = \frac{1}{2} \sqrt{d {}^{2} + e {}^{2} - 4f}$

$r = \frac{1}{2} \sqrt{0 {}^{2} + 0 {}^{2} - 4( - 12)}$

$r = \frac{1}{2} \sqrt{48 }$

$r = 2 \sqrt{3}$

$x {}^{2} + y {}^{2} - 16 = 0$

$c = ( \frac{ - 0}{2} ; \frac{ - 0}{2} )$

$r = \frac{1}{2} \sqrt{0 {}^{2} + 0 {}^{2} - 4( - 16) }$

$r = \frac{1}{2} \sqrt{64}$

$r = 4$

$x {}^{2} + y {}^{2} - 4x + 6y = 0$

$c = ( \frac{ - ( - 4)}{2} ; \frac{ - 6}{2} )$

$c = 2; - 3$

$r = \frac{1}{2} \sqrt{( - 4) {}^{2} + (6) {}^{2} - 4(0) }$

$r = \frac{1}{2} \sqrt{16 + 36}$

$r = \frac{1}{2} \sqrt{52}$

$r = \sqrt{13}$

$x {}^{2} + y {}^{2} - 10x + 2y + 22 = 0$

$c = ( \frac{ - ( - 10)}{2} ; \frac{ - 2}{2} )$

$c = 5; - 1$

$r = \frac{1}{2} \sqrt{( - 10) {}^{2} + (2) {}^{2} - 4(22) }$

$r = \frac{1}{2} \sqrt{100 + 4 - 88}$

$r = \frac{1}{2} \sqrt{16}$

$r = 2$

en la 2 ecuacion el c=(0;0)