Question

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

Answer 1

Answer:

$(x+6)^2+(y-5)^2=41$

Step-by-step explanation:

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

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

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

$r = \sqrt{(-6 + 1)^2 + (5 - 1)^2} = \sqrt{5^2+4^2}=\sqrt{41}\\\\r^2 = 41$

So we can obtain the equation $(x+6)^2+(y-5)^2=41$