michelleruiz723 michelleruiz723
  • 25-03-2020
  • Mathematics
contestada

Find an equation of the circle that has center (- 6, 5) and passes through (- 1, 1) .

Respuesta :

kimchiboy03 kimchiboy03
  • 25-03-2020

Answer:

[tex](x+6)^2+(y-5)^2=41[/tex]

Step-by-step explanation:

The equation of a circle follows the general equation [tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Since the center of the circle is (-6, 5), we progress to the formula equation [tex](x + 6)^2 + (y- 5)^2 = r^2[/tex]

Since this is a circle, the distance between (-6, 5) and (-1, 1) is the radius of the circle.

[tex]r = \sqrt{(-6 + 1)^2 + (5 - 1)^2} = \sqrt{5^2+4^2}=\sqrt{41}\\\\r^2 = 41[/tex]

So we can obtain the equation [tex](x+6)^2+(y-5)^2=41[/tex]

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