Which is an equation of a circle with center (-5, -7) that passes through the point (0, 0)?

(x + 5)2 + (y + 7)2 = 74
(x - 5)2 + (y - 7)2 = 74
(x - 5)2 + (y - 7)2 = 37
(x + 5)2 + (y + 7)2 = 37

Question

Which is an equation of a circle with center (-5, -7) that passes through the point (0, 0)?
 	

	(x + 5)2 + (y + 7)2 = 74
	(x - 5)2 + (y - 7)2 = 74
	(x - 5)2 + (y - 7)2 = 37
	(x + 5)2 + (y + 7)2 = 37

anonymous · Answer

(x-k)^+(y-h)^2=r2

anonymous · Answer

k is x and h is y and those are the centers

mathmale · Answer

First, would you look up (if necessary) and type out here the standard equation of a circle with center at (h,k) and radius r?  joshfers is on the right track, but is not fully correct.  Note that I'm addressing you, @fack44.

anonymous · Answer

(x - h)2  +  (y-k)2   =   r2

anonymous · Answer

right.I forgot the square on the first equation and put inverse the letters.Now just plug the centers that they gave you in the equation.

mathmale · Answer

sorry to be picky, but the SQUARE of (x-h) needs to be written properly:

(x-h)^2 or $$(x-h)^2$$    .... but not as (x-h)2.  Or you could use the Draw utility (below):|dw:1403186610015:dw|

mathmale · Answer

Please write or type the standard equation of a circle with center at (h,k) and radius r.

anonymous · Answer

so is it (x - 5)^2 + (y - 7)^2= 37

anonymous · Answer

@mathmale

mathmale · Answer

I was hoping you'd write the standard equation of a circle with center at (h,k) and radius r:$$(x-h)^2+(y-k)^2 = r^2$$

mathmale · Answer

The center of your circle is  (-5, -7).   Thus, your h = -5 and your y = -7.  substitute these into the standard equation (above).