Question

The circle below is centered at the point (4,-3) and has a radius of length 3. What is its equation? 6 O A. (x - 4)2 + (y + 3)2 - 32 W O B. (x-3)2 + (y + 4)2 = 9 ОООО C. (X- 3)2 + (y - 4)2 = 9 D. (x + 4)2 + (y - 3)2 = 32. ​

Answer 1

Answer:

$(x-4)^2+(y+3)^2=9$

Step-by-step explanation:

Standard Form equation of a circle with center (h, k) and radius r:

$(x-h)^2+(y-k)^2=r^2$

Plug in given values: h = 4, k = -3, r = 3.

$(x-4)^2+[y-(-3)]^2=3^2\\\\(x-4)^2+(y+3)^2=9$

The tricky part can be the sign in the y squard term:  y - (-3) = y + 3.