Write the equation of the circle given the following conditions center (7,-4) radius 6. Can someone help explain how to do this I have NO idea how to.
a. (x-7)^2 + (y+4)^2 = 36
b. (x+7)^2 + (y+4)^2 = 36
c. (x-7)^2 + (y+4)^2 = 6
d. (x-7)^2 + (y-4)^2 = 36

Question

Write the equation of the circle given the following conditions  center (7,-4) radius 6. Can someone help explain how to do this I have NO idea how to.
a. (x-7)^2 + (y+4)^2 = 36
b. (x+7)^2 + (y+4)^2 = 36
c. (x-7)^2 + (y+4)^2 = 6
d. (x-7)^2 + (y-4)^2 = 36

blockcolder · Answer

If you're given a center (h,k) and a radius r, simply substitute h, k, and r into this equation:
$$(x-h)^2+(y-k)^2=r^2$$
In your case, h=7, k=-4, r=6.

anonymous · Answer

Standard form of the equation of a circle with center (h,k) and radius r is$$(x-h)^2+(y-k)^2=r^2$$This comes from the Pythagorean theorem.  If you pick a point on the circle (suppose its coordinates are (x,y)), draw the radius from the center (h,k) to that point.  In all except four cases, you can make a right triangle with the legs horizontal and vertical.  The length of these legs are |x-h| and |y-k|.  Put those into the Pythagorean theorem, and you get the equation of the circle above.  This works for any point on the circle, including the four points that make no triangle.