Question

Can’t figure this out, really need help

Answer 1

The equation of the graphed circle is:

$(x+3)^2+(y-6)^2=13$

The equation of a circle with center (a, b) is:

$(x - a)^2+(y-b)^2=r^2$

The circle shown passes through the point (0, 8) and has a center (-3, 6)

The radius of the circle is calculated as:

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\r=\sqrt{(0-(-3))^2+(8-6)^2}\\\\r=\sqrt{9+4} \\\\r=\sqrt{13}$

Substituting a = -3, b = 6, and $r=\sqrt{13}$ into the equation $(x - a)^2+(y-b)^2=r^2$

$(x-(-3))^2+(y-6)^2=\sqrt{13^2} \\\\(x+3)^2+(y-6)^2=13$

Therefore, the equation of the circle is:

$(x+3)^2+(y-6)^2=13$

Learn more on equation of a circle here: brainly.com/question/17064049