Convert each equation to its polar form.x2+(y−2)2=4x^2+(y-2)^2=4x2+(y−2)2=4

r = 4 sin ⁡ θ r=4\sin\theta r = 4 sin θ

Convert each equation to its polar form. x 2 + ( y − 2 ) 2 = 4 x^2+(y-2)^2=4 x 2 + ( y − 2 ) 2 = 4

In this problem, we're asked to convert the equation into its polar form. And the equation that we're given is x squared plus y minus 2 squared is equal to 4. So let's go ahead and dive in. Now remember that we want to replace x, y, and x squared plus y squared using these formulas. Now here you might be tempted to try to replace this using that r squared value. But since we have y minus 2 being squared and not just y, we can't do that. So from here, we actually have to replace our squared x and y values using our cosine theta and r sine theta. So if I replace x with r cosine theta, y with r sine theta, that gives me r cosine theta, and that whole term is squared, plus r sine theta minus 2, and all of that is squared. Then on that right side, I still just have a 4. Now we're going to have to do a bit more algebra than we have for our other practice problems but that's fine. Make sure that you take your time here because it can be easy to just miss a step in your algebra. Now I want to go ahead and get all of these values squared. Remember that if you have an r cosine theta being squared remember to square that r. So squaring that whole term will give me r squared times the cosine squared of theta. Now here, since I have r sine of theta minus 2, this ends up giving me more than just one term. So if I go ahead and square these, this will give me r squared times the sine squared of theta minus 4 r sine theta and then a plus 4. Then all of this is equal to 4. Now from here, let's see if we can do anything here to cancel some stuff out. Now I see that I have this 4 and also this 4 on the right. So if I were to subtract 4 from both sides, those fours are going to cancel. Now I have r squared times the cosine squared of theta plus r squared times the sine squared of theta minus 4 r sine theta is equal to 0. Now where do I go from here? Because this looks like a lot is going on and we still wanna try to solve for r. Now remember that you want to be constantly scanning for and recognizing if there are trig identities. Now here one thing that I notice is that I have a cosine squared theta and a sine squared theta. So let's go ahead and factor out our r squared term from these. This gives me r squared times cosine squared of theta plus the sine squared of theta, and then still having that term on the end minus 4 r sine theta, and all of that is equal to 0. Now based on my Pythagorean Shirk identities, I know that this is just equal to 1 so this entire term is just r squared because r squared times 1 is just r squared. Now I have r squared minus 4 r sine theta So this is looking much more simple. So let's go ahead and do a bit more algebra. Now I'm gonna go ahead and add 4 r sine theta to both sides. It will cancel on that left side, leaving me with r squared, and that's equal to 4 r sine theta. Now since I have an r and an r squared one of those will cancel out and my final equation that I'm left with here is r is equal to 4 times the sine of theta and this is my equation in its polar form. Thanks for watching and let me know if you have any questions.