# What is the relationship between variable magnitude and a circle-tangent vector field?

The following vector field $$\vec G(x,y) = \dfrac{-y\hat \imath + x\hat \jmath}{\sqrt{x^2+y^2}}$$ shows vectors that are tangent to circle centered at the origin. However, the text I am using also mentions that the magnitudes of these vectors are equal to their distances from the origin.

My question is, wouldn't you have to mutliply both components by $\sqrt{x^2+y^2}$ rather than divide?

I now see that $|\vec G| = 1$, something I should have noticed before. My new approach to get the magnitudes to equal to the distances of the points from the origin:

$$|\vec G_1| = \sqrt{k^2\cdot(y^2 + x^2)} = \sqrt{y^2+x^2} \implies k=1 \implies \vec G_1(x,y) = -y\hat \imath + x\hat \jmath$$

So is the text (as I have written it) simply incorrect or is there something else behind it?

• I may disagree with your text. The length of $G$ is constantly 1. – Randall Sep 18 '17 at 1:27
• Do you see any other reason for this? I have edited the question. – Rithwik Sudharsan Sep 18 '17 at 2:34
• The text would be correct without the denominator: the length of the vector $-yi + xj$ is in fact equal to its distance from the origin. – Randall Sep 18 '17 at 19:48
• Thanks! I speculated so, I must have read the text wrong if it's not incorrect. – Rithwik Sudharsan Sep 21 '17 at 4:59