Given: a circle is given with center (3,-3) and equation
[tex](x-B_1)^2+(y-B_2)^2=(B_3)^2[/tex]
Find:
[tex]B_{1,\text{ }}B_{2,}B_3[/tex]
Explanation: the general equation of the circle with center (a,b) and radius r is
[tex](x-a)^2+(y-b)^2=r^2[/tex]
in the given figure the center of the circle is at (3,-3)
so the equatio of the circle becomes
[tex](x-3)^2+(y+3)^2=(3)^2[/tex]
on comparing eith the given equation we get
[tex]B_1=3,\text{ B}_2=-3\text{ and B}_3=3[/tex]
