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Sagot :
Let A = (5,-3) and B = (7,0). Notice that point A is the center of the circle, then:
[tex](h,k)=(5,-3)=A[/tex]now, point B is on the circumference of the circle, then, the distance between A and B is the radius of the circle:
[tex]\begin{gathered} r=d(A,B)=\sqrt[]{(5-7)^2+(-3-0)^2}=\sqrt[]{(-2)^2+(-3)^2}=\sqrt[]{4+9}=\sqrt[\square]{13} \\ \Rightarrow r=\sqrt[]{13} \end{gathered}[/tex]then, using the equation of the circle with center (h,k) and radius r:
[tex](x-h)^2+(y-k)^2=r^2[/tex]in this case, we have the following:
[tex]\begin{gathered} (h,k)=(5,-3) \\ r=\sqrt[]{13} \\ \Rightarrow(x-5)^2+(y-(-3))^2=(\sqrt[]{13})^2 \\ \Rightarrow(x-5)^2+(y+3)^2=13 \end{gathered}[/tex]therefore ,the equation of the circle is (x-5)^2 + (y+3)^2 = 13
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