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

Question

Aaliyah Abdullah · Accepted Answer

#r^2-8rcostheta-10rsintheta+37=0# Relation between polar coordinates #(r,theta)# and rectangular coordinates is given by #x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#. Hence #4=(x-4)^2+(y-5)^2# can be written as #x^2-8x+16+y^2-10y+25=4# or #x^2+y^2-8x-10y+37=0# or #r^2-8rcostheta-10rsintheta+37=0#

Grace Ashcraft · Answer

undefined To convert the equation $4 = (x - 4)^2 + (y - 5)^2$ into polar form, we can follow these steps:

1. Replace $x$ and $y$ with their respective polar coordinate representations:
   $$
   x = r\cos(\theta) \quad \text{and} \quad y = r\sin(\theta)
   $$

2. Substitute these expressions into the given equation:
   $$
   4 = (r\cos(\theta) - 4)^2 + (r\sin(\theta) - 5)^2
   $$

3. Expand and simplify the equation:
   $$
   4 = r^2\cos^2(\theta) - 8r\cos(\theta) + 16 + r^2\sin^2(\theta) - 10r\sin(\theta) + 25
   $$

4. Combine like terms and rearrange:
   $$
   4 = r^2(\cos^2(\theta) + \sin^2(\theta)) - 8r\cos(\theta) - 10r\sin(\theta) + 41
   $$

5. Recognize that $\cos^2(\theta) + \sin^2(\theta) = 1$ (the Pythagorean identity), and substitute:
   $$
   4 = r^2 - 8r\cos(\theta) - 10r\sin(\theta) + 41
   $$

6. Rearrange to isolate $r$:
   $$
   r^2 - 8r\cos(\theta) - 10r\sin(\theta) = -37
   $$

7. Complete the square for $r$:
   $$
   r^2 - 8r\cos(\theta) + 16\cos^2(\theta) + 10r\sin(\theta) + 25\sin^2(\theta) = -37 + 16\cos^2(\theta) + 25\sin^2(\theta)
   $$

8. Factor the perfect square trinomials:
   $$
   (r - 4\cos(\theta))^2 + (5\sin(\theta))^2 = 4
   $$

9. Divide both sides by 4 to isolate $r$:
   $$
   \frac{(r - 4\cos(\theta))^2}{4} + \frac{(5\sin(\theta))^2}{4} = 1
   $$

10. Recognize that $r^2 = x^2 + y^2$ and $a^2 = (r - h)^2$ and $b^2 = (r - k)^2$ (where $h$ and $k$ are the coordinates of the center of the ellipse) to rewrite the equation:
   $$
   \frac{(x - 4)^2}{2^2} + \frac{(y - 0)^2}{(\frac{5}{2})^2} = 1
   $$

Thus, the equation $4 = (x - 4)^2 + (y - 5)^2$ in polar form is $\frac{(x - 4)^2}{2^2} + \frac{(y - 0)^2}{(\frac{5}{2})^2} = 1$.