Question

Determine the equation of the circle graphed below. 12 11 10 9 8 7 5 1 (7,0) 3 4 5 6 7 8 9 10 11 12 -12-11-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 (5,-3). -6 -8 -10 -12

Answer 1

Let A = (5,-3) and B = (7,0). Notice that point A is the center of the circle, then:

`(h,k)=(5,-3)=A`

now, point B is on the circumference of the circle, then, the distance between A and B is the radius of the circle:

`\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}`

then, using the equation of the circle with center (h,k) and radius r:

`(x-h)^2+(y-k)^2=r^2`

in this case, we have the following:

`\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}`

therefore ,the equation of the circle is (x-5)^2 + (y+3)^2 = 13