Locus of Midpoints of Chords in Circle x^2+y^2-2y-3=0

[Mentallic]

Homework Statement

Find the equation of the locus of midpoints of all chords of length 2 units in the circle with equation $$x^2+y^2-2y-3=0$$

Homework Equations

$$d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$$

The Attempt at a Solution

I don't know how to begin solving this problem. All I know is $$(x_2-x_1)^2+(y_2-y_1)^2=4$$, and the variables, $$(x_1,y_1)$$ and $$(x_2,y_2)$$ satisfy the circle equation.

I transformed the circle equation into the general form ~ $$x^2+(y-1)^2=4$$ So the circle is centred $$(0,1)$$ and radius 2.
Actually while writing this, I realize the locus of the circle will have the same centre thus, $$x^2+(y-1)^2=r^2$$, and the perpendicular bisector of a chord in a circle passes through its centre, so I can use pythagoras' theorem:

$$c^2=r^2+b^2$$

$$4=r^2+(\frac{2}{2})^2$$

$$r^2=3, r=\sqrt{3}$$

Therefore, the circle equation is: $$x^2+(y-1)^2=3$$

Somehow while trying to explain my problem, I figure it out for myself? Anyway, are there any other methods to solve this problem? I thought it would've involved the distance formula in some way.

 

[Tedjn]

You are using the distance formula when you use Pythagoras' theorem. But what makes you think that you would need the distance formula? The trick behind most locus questions is first figuring out what the locus should look like. In this case, it was a circle. Then, justify it, as you have done with Pythagoras' theorem.

 

[Mentallic]

Yeah, but without realizing it would be another circle with the same centre, I wouldn't have gotten very far.

 

[Tedjn]

That is normal with locus problems. You often need to plot part of it in order to get an idea of where to start.

 

[Mentallic]

I guess I'm just a little skeptical about it because I've been working on locii in the complex plane too, and since I'm new to the whole idea of complex numbers and graphing them, I haven't been able to even visualize what I should get, but I still get the answer nonetheless (eventually).