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Question: How do you convert\({(x - 1)^2} + {y^2} = 5\)in polar form?...

How do you convert(x−1)2+y2=5{(x - 1)^2} + {y^2} = 5in polar form?

Explanation

Solution

Hint : The polar form of a curve or a point is different from the conventional nomenclature that we use for denoting a curve or a point on the coordinate plane . The normal (x,y)(x,y)that we use is called the Cartesian form or the rectangular form. The polar form whereas is not written in that form it is written as the function of the distance of a point from the centre (r)(r) and the angle the given point makes from the positivexxaxis (θ)(\theta )
The polar form for the above given curve can be found out by substituting xxand yyin the given equation with the formula
x=rcos⁡θx = r\cos \theta
And y=rsin⁡θy = r\sin \theta

Complete step-by-step answer :
Given,
(x−1)2+y2=5{(x - 1)^2} + {y^2} = 5
We substitute standard formula for converting the given Cartesian form into the polar form which is given as:
x=rcos⁡θx = r\cos \theta
And y=rsin⁡θy = r\sin \theta
The value of the given curve will then be
(rcosθ−1)2+(rsinθ)2=5{(rcos\theta - 1)^2} + {(rsin\theta )^2} = 5
Upon further solving we will get
r2cos2θ−2rcosθ+1+r2sin2θ=5{r^2}co{s^2}\theta - 2rcos\theta + 1 + {r^2}si{n^2}\theta = 5
Now we will factor out r2{r^2}out of this and
r2(cos2θ+sin2θ)−2rcosθ+1−5=0{r^2}(co{s^2}\theta + si{n^2}\theta ) - 2rcos\theta + 1 - 5 = 0
Since cos2θ+sin2θco{s^2}\theta + si{n^2}\theta =11
We will write
r2−2rcosθ−4=0{r^2} - 2rcos\theta - 4 = 0
Upon solving for the above quadratic equation using the quadratic equation formula we get,
r=−(−2cosθ)±(−2cosθ)2−4⋅1⋅(−4)2⋅1r = \dfrac{{ - ( - 2cos\theta ) \pm \sqrt {{{( - 2cos\theta )}^2} - 4 \cdot 1 \cdot ( - 4)} }}{{2 \cdot 1}}
And we can write
r=(2cosθ)±4cos2θ+162r = \dfrac{{(2cos\theta ) \pm \sqrt {4co{s^2}\theta + 16} }}{2}
Thus in the end we can elegantly write
r=cosθ±cos2θ+4r = cos\theta \pm \sqrt {co{s^2}\theta + 4}
Thus r=cosθ±cos2θ+4r = cos\theta \pm \sqrt {co{s^2}\theta + 4} is the polar form of the above given curve.
So, the correct answer is “r=cosθ±cos2θ+4r = cos\theta \pm \sqrt {co{s^2}\theta + 4} ”.

Note : The given polar form of the point or the curve can also be easily converted back to the starting equation using the same standard equations given earlier
x=rcos⁡θx = r\cos \theta
And y=rsin⁡θy = r\sin \theta
Thus these forms of denoting a point or a curve are interchangeable between the polar and the Cartesian coordinates(also called the rectangular coordinates) .