Respuesta :
Answer:
the region of the circle in polar coordinate can be expressed as:
[tex]\mathbf{- \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2} , 0 \leq r \leq 11 cos \theta}[/tex]
Step-by-step explanation:
Given that:
[tex]x^2 +y^2 = 11x[/tex]
This implies that :
[tex](x-5.5)^2 + y^2 = 5.5^2[/tex]
center (5.5, 0) and radius = 5.5
The relation between the cartesian coordinates (x,y) and the polar coordinates (r, θ) is as follows:
x = rcosθ
y = rsinθ
Thus, [tex]x^2 + y^2 = r^2[/tex]
[tex]x^2 + y^2 = 11x[/tex]
[tex](rcos \theta)^2 + (r sin \theta)^2 = 11(rcos \theta)[/tex]
[tex]r^2 = 11r cos \theta[/tex]
Thus, the angle θ runs between [tex]- \dfrac{\pi}{2}[/tex] to [tex]\dfrac{\pi}{2}[/tex]
Therefore, the region of the circle in polar coordinate can be expressed as:
[tex]\mathbf{- \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2} , 0 \leq r \leq 11 cos \theta}[/tex]
The required region is,
[tex]x=\frac{11}{2}+\frac{11}{2}cos\theta , 0\leq \theta\leq2\pi \\y=\frac{11}{2}sin\theta ,0\leq r \leq \frac{11}{2}[/tex]
Given circle is,
[tex]x^2 +y^2 = 11x[/tex]
Computation:
Solving the given circle,
[tex]x^2-11x+y^2=0\\x^2-2(\frac{11x}{2})+y^2=0[/tex]
Adding [tex]\frac{11}{2}[/tex] both sides we get,
[tex]x^2-2(\frac{11x}{2})+\frac{121}{4}+y^2=\frac{121}{4}\\(x-\frac{11}{2})^2+y^2=(\frac{11}{2})^2[/tex]
It represents a circle having the radius [tex]\frac{11}{2}[/tex] and centred at [tex](\frac{11}{2},0)[/tex]
So, in polar coordinate,
[tex]x=\frac{11}{2}+\frac{11}{2}cos\theta , 0\leq \theta\leq2\pi \\y=\frac{11}{2}sin\theta ,0\leq r \leq \frac{11}{2}[/tex]
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