Question 1039381
.
Write the equation in standard form of the circle whose center is (3,-5) and passes through point (7,1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


The equation of the circle is 


{{{(x-3)^2 + (y-(-5))^2}}} = {{{r^2}}},


where "r" is the radius, or


{{{(x-3)^2 + (y+5)^2}}} = {{{r^2}}}.


"r" is the distance from the center to the given point in the circle, which is


{{{r^2}}} = {{{(7-3)^2 + (1 - (-5))^2}}} = {{{4^2 + 6^2}}} = 16 + 36 = 52.


Finally, the equation of the circle is 


{{{(x-3)^2 + (y+5)^2}}} = {{{52}}}.