Question

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

Answer 1

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 $(x,y)$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 $(r)$ and the angle the given point makes from the positive$x$axis $(\theta )$
The polar form for the above given curve can be found out by substituting $x$and $y$in the given equation with the formula
  $x = r\cos \theta $
And $y = r\sin \theta $

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

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 = r\cos \theta $
And $y = 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) .