How do you convert #9=(x+5)^2+(y+4)^2# into polar form?

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Answer 1 · 2016-10-02T23:11:54

#r = -(10cos(theta) + 8sin(theta))/2 + sqrt((10cos(theta) + 8sin(theta))² - 128)/2#

Expand the squares using the pattern #(a + b)² = a² + 2ab + b²#

#9 = x² + 10x + 25 + y² + 8y + 16#

Substitute the 3 following equations into the above where appropriate:

#x² + y² = r²#
#x = rcos(theta)#
#y = rsin(theta)#

#9 = r² + 10rcos(theta) + 8rsin(theta) + 41#

Write as a quadratic equation in r:

#0 = r² + (10cos(theta) + 8sin(theta))r + 32#

Use the positive root of the quadratic formula to solve for r:

#r = -b/(2a) + sqrt(b² - 4(a)(c))/(2a)#

#r = -(10cos(theta) + 8sin(theta))/2 + sqrt((10cos(theta) + 8sin(theta))² - 128)/2#