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Sagot :

The equation of the graphed circle is:

[tex](x+3)^2+(y-6)^2=13[/tex]

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

[tex](x - a)^2+(y-b)^2=r^2[/tex]

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

The radius of the circle is calculated as:

[tex]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}[/tex]

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

[tex](x-(-3))^2+(y-6)^2=\sqrt{13^2} \\\\(x+3)^2+(y-6)^2=13[/tex]

Therefore, the equation of the circle is:

[tex](x+3)^2+(y-6)^2=13[/tex]

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