How do you convert 1=(x+4)^2+(y+7)^2 into polar form?
1 Answer
Apr 2, 2016
Explanation:
Let
1 = (x + 4)^2 + (y + 7)^2
1 = (r cos(theta) + 4)^2 + (r sin(theta) + 7)^2
1 = (r^2 cos^2(theta) + 8 r cos(theta) + 16)
+ (r^2 sin^2(theta) + 14 r sin(theta) + 49)
1 = r^2 (cos^2(theta) + sin^2(theta)) + r (8 cos(theta) + 14 sin(theta)) + 65
0 = r^2 + 2r (4 cos(theta) + 7 sin(theta)) + 64
Complete the square
(r + (4 cos(theta) + 7 sin(theta)))^2 + 64 = (4 cos(theta) + 7 sin(theta))^2
r = +-sqrt{(4 cos(theta) + 7 sin(theta))^2-64} - (4 cos(theta) + 7 sin(theta))
