a) We first know that the equation #(x-a)^2+(y-b)^2=r^2# is the equation for a circle where #a# is the #x# coordinate of the center of the circle, #b# is the #y# coordinate of the center of the circle, and #r# is the radius of the circle.
We can find the center coordinate of the circle by adding the x and y coordinates of #A# and #B# against each other and then dividing by #2#. For the #x# coordinate, we do #(3+7)/2# to get #5# as the #x# coordinate for the center. We now do the same for the #y# coordinate to get #(4-6)/2=-1#.
We now replace #a# and #b# to now get the equation #(x-5)^2+(y+1)^2=r^2#. Now all we have to do is find #r#. We use the Pythagorean Theorem to find that the length of #AB# is #sqrt((7-3)^2+(4+6)^2)=sqrt(116)=2sqrt(29)#. If we know that the diameter is #2sqrt(29)#, then the radius is #sqrt(29)#. We replace #r# with 29 to get #(x-5)^2+(y+1)^2=29#.
b) We name the center of the new circle #C#. The center of the old circle should be named #D#. We know that #A# must be the center of #CD#. We find that the center of the new circle is #( 1 , 9 )# and because the radius is the same, the new equation is #(x-5)^2+(y+1)^2=29#.
