A circle is centered at the point (5, -4) and passes through the point (-3, 2).The equation of this circle is

Answer:[tex](x-5)^{2} +(y+4)^{2} =10^{2}[/tex]Step-by-step explanation:To calculate the formula of a circle that is not in the center fo the graph, you need two things, the point where the cirlce is centered and any point in the circumference, and with this you can calculate the radius, to calculate the first part you just need the next formula:[tex](x-x^{c})+(y-y^{c})=r^{2}[/tex]Where [tex]x^{c}y^{c}[/tex] are the x and y where the center of the circle is, so you just evaluate with the values you are given:[tex](x-x^{c})+(y-y^{c})=r^{2}[/tex][tex](x-(5))+(y-(-4))=r^{2}[/tex][tex](x-5)+(y+4)=r^{2}[/tex]Now that you have the first part, we can calculate the radius, remember taht the radius of any given circumference in the graph is the hypotenuse of the X´s and Y´s that are part of those two points given, so we just calculate it like any hypotenuse:[tex]c=\sqrt{x^{2}+ y^{2} }[/tex]To calculate this we would rest to the point in the circumference, the center of the circumference, like this:[tex]c=\sqrt{(x^{2}-x^{1})^{2}+ (y^{2}-y^{1})^{2} }[/tex][tex]c=\sqrt{(-3-5)^{2}+ (2-(-4))^{2} }[/tex][tex]c=\sqrt{(-8)^{2}+ (6)^{2} }[/tex][tex]c=\sqrt{64+ 36 }[/tex][tex]c=\sqrt{100 }[/tex][tex]c=10[/tex]So your radius would be 10, now we just put that into our previous formula:[tex](x-5)+(y+4)=r^{2}[/tex][tex](x-5)+(y+4)=10^{2}[/tex]So the formula for the circle that is centered at (5,-4) and passes through the point (-3,2) would be:[tex](x-5)+(y+4)=10^{2}[/tex]