Cartesian to Polar Coordinates

The radius comes straight from Pythagoras, but the angle needs care - why the inverse tangent alone is not enough, and the three cases decided by the sign of x.

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Cartesian to Polar Coordinates — Moosa Academy

Going from polar to Cartesian is direct: two formulas and you are done. Going the other way is not quite so simple. The radius is easy, but the angle needs care — because a formula alone cannot tell which side of the plane the point is on.

Concept Finding the radius

The values

x

and

y

are the two legs of a right triangle, and

r

is its hypotenuse. So the radius comes straight from Pythagoras.

r = \sqrt{x^2 + y^2}

This one rule has no exceptions. It works for positive and negative values alike, and the answer is always positive.

Concept Why the angle is harder

The starting rule for the angle is:

\theta = \tan^{-1}\!\left( \frac{y}{x} \right)

But this is incomplete. Two points on opposite sides of the plane can give the same ratio $y/x$ :

For

(1, 1)

: first quadrant,

\theta = 45^\circ

, and

y/x = 1/1 = 1

For

(-1, -1)

: third quadrant,

\theta = 225^\circ

, and

y/x = (-1)/(-1) = 1

⟹ same ratio, opposite directions

The division throws away the signs, so the calculator cannot tell the two points apart. The fix is to check the sign of $x$ first, and let that decide what to do.

Theorem The three cases for θ

If $x > 0$ — the point is on the right-hand side, so the formula is already correct:

\theta = \tan^{-1}(y/x)

If $x < 0$ — the point is on the left-hand side, and the formula lands in the wrong half. Add a half turn:

\theta = \tan^{-1}(y/x) + 180^\circ

If $x = 0$ — the point sits on the vertical axis and

y/x

is undefined. Read the angle off the sign of

y

instead:

y > 0 \Rightarrow \theta = 90^\circ

y < 0 \Rightarrow \theta = -90^\circ

Always look at $x$ before reaching for the calculator. That single check decides which of the three routes to take.

Example A point with x positive

Convert $(2, 3)$ to polar form.

Step 1 — the radius:

r = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61

Step 2 — the sign of

x

:

x = 2 > 0

, so use the formula directly

Step 3 — the angle:

\theta = \tan^{-1}(3/2) = \tan^{-1}(1.5) \approx 56.31^\circ

⟹

(\sqrt{13},\; 56.31^\circ)

Example A point with x negative

Convert $(-2, 3)$ to polar form.

Step 1 — the radius:

r = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61

Step 2 — the sign of

x

:

x = -2 < 0

, so a half turn must be added

Step 3 — the raw value:

\tan^{-1}(3/(-2)) = \tan^{-1}(-1.5) \approx -56.31^\circ

Step 4 — correct it:

-56.31^\circ + 180^\circ = 123.69^\circ

⟹

(\sqrt{13},\; 123.69^\circ)

Notice the radius is the same as the first example — squaring erases the minus sign. Only the angle tells the two points apart.

Example A point on the axis

Convert $(0, 5)$ to polar form.

Step 1 — the radius:

r = \sqrt{0^2 + 5^2} = \sqrt{25} = 5

Step 2 — the sign of

x

:

x = 0

, so this is the special case

Step 3 — the sign of

y

:

y = 5 > 0

, so

\theta = 90^\circ

⟹

(5,\; 90^\circ)

No inverse tangent is needed here at all. The point is straight up the vertical axis, and the angle is read off directly.

Summary

The radius is always r = √(x² + y²), with no exceptions.
The angle starts from θ = tan⁻¹(y/x), but that alone is not enough.
If x > 0, the formula is correct as it stands.
If x < 0, add 180° to the result.
If x = 0, use the sign of y: positive gives 90°, negative gives −90°.
Always check the sign of x first — it decides which route to take.

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