How do you graph the system of polar equations to solve $r = 1 + cos θ$ $r=1+costheta$ and $r = 1 - cos θ$ $r=1-costheta$ ?

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Answer 1 · 2016-11-14T02:27:39

graph{x^2+y^2-sqrt(x^2+y^2)-x=0 [-10, 10, -5, 5]} Common points are at $(r, θ) = (1, \pm \frac{π}{2})$ $(r, theta)=(1, +-pi/2)$

graph{(x^2+y^2+x)^2-x^2-y^2=0 [-10, 10, -5, 5]}

Using the cartesian equivalents $(x^{2} + y^{2} \pm x) - \sqrt{x^{2} + y^{2}} = 0$ $(x^2+y^2+-x)-sqrt(x^2+y^2)=0$

for $r = 1 \pm cos θ$ $r=1+-cos theta$ , the graphs are obtained. The graphs are

are symmetrically opposite, about the y-axis .

By this symmetry, the two meet on the vertical axis, given in

halves, by $θ = \pm \frac{π}{2}$ $theta=+-pi/2$ , wherein r=1, at both the common points.

Algebraically adding and subtracting $r = 1 \pm cos θ$ $r=1+-cos theta$ , the solutions

$(r, θ) = (1, \pm \frac{π}{2})$ $(r, theta)=(1, +-pi/2)$ can be obtained.we y

How do you graph the system of polar equations to solve r=1+cosθr=1+costheta and r=1−cosθr=1-costheta?