Calculus (3rd Edition)

$(x - 2)^{2}$ + $(y-4)^{2}$ = 9 $(x - 2)^{2}$ + $(y-4)^{2}$ = 26
$(x - 2)^{2}$ + $(y-4)^{2}$ = $r^{2}$ gives a circle centered at $(2,4)$ with radius r, using the general form equation for circles a) plugging in r = 3 gives $(x - 2)^{2}$ + $(y-4)^{2}$ = 9 b) the circle at $(2,4)$ passes through $(1,-1)$ if it has radius equal to the distance between the two points, so r = $\sqrt ((2-1)^{2} + (4 - (-1))^{2})$ = $\sqrt (1+25)$ = $\sqrt 26$ Plugging into the general equation gives $(x - 2)^{2} + (y-4)^{2}$ = 26