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I have no clue, I know that the radius is 2, will that mean that i have to go 2 units up every time?

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    What is that $(1, -\sqrt{3})$ floating there? + Can you upload a picture with the complete text? – 2017-02-07
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    I think that $(1,-\sqrt{3})$ is correcting $D=(1,-3)$ that corrects the radius $OD$ from $\sqrt{10}$ to $OC=2$. – 2017-02-07
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    No wonder you're confused! It's a very poorly written problem. – 2017-02-07

2 Answers 2

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Since the given radius of $r=2$ conflicts with the given coordinates of $D=(1,-3)$ we may assume that the insertion of the coordinates $(1,-\sqrt{3})$ was intended as a correction of a typographical error in the original.

Since the coordinates of a point $P=(x,y)$ on a circle with equation $x^2+y^2=r^2$ are given by $x=r\cos\theta,\,y=r\sin\theta$, the coordinates of the point $C=(2\cos40^\circ,2\sin40^\circ)$

In general the coordinates of a point on a circle centered at $(0,0)$ and having radius $r$ are equal to $(r\cos\theta,r\sin\theta)$ where $\theta$ is the angle between the segment connecting the center to the point and the positive $x$-axis as depicted in the following diagram.

Using the trigonometric identity $\cos^2\theta+\sin^2\theta=1$ we see that the sum of the squares of the coordinates is

$$ x^2+y^2=(r\cos\theta)^2+(r\sin\theta)^2=r^2(\cos^2\theta+\sin^2\theta)=r^2\cdot1=r^2$$

Therefore the coordinates satisfy the equation

$$ x^2+y^2=r^2 $$

coordinates of point on circle

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    how did u get (2cos40,2sin40)? – 2017-02-16
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    On a circle with radius $r$ centered at the origin $O$ the coordinated of a point $C$ on the circle are $(r\cos\theta,r\sin\theta)$ where $\theta$ is the angle the line segment $OC$ makes with the positive $x$-axis. Give me a few minutes and I will add a diagram illustrating this fact. – 2017-02-16
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I have no idea why the radius is given as $2$ when we can clearly see that the point $D $ also lies on it.

We know that the parametric equation of a circle $x^2 + y^2 =a^2$ is given by $(a\cos \theta, a\sin \theta) $.

Substituting the point $(1, -3) $ in the equation of the circle, we get its equation as $x^2+y^2=10$. Thus the parametrization of a point on it is given by $(\sqrt{10}\cos \theta, \sqrt {10}\sin \theta) $.

You can find the angle $X $ (in your diagram) by equating the point and the parametric form. Hope it helps.

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    Perhaps the insertion of the coordinates $(1,-\sqrt{2})$ in a different font was intended as a correction for the coordinates of $D$. – 2017-02-07