Find locus of center of circle passing origin, A and B, with AB at a constant distance $c$ from origin and intersecting axes at A and B.

Question

Find the locus of the circle passing through O, A and B.

Let the line intersect X and Y axis at (2h,0) and (0,2k) respectively.

The line is the diameter for the given circle, so the centre will be (h,k).

The radius of the circle is $c$

Then $$h^2+k^2=c^2$$

This is the locus for the centre of the circle. However, i am not able to find the locus for the entire circle itself, because the centre isn’t clearly defined.

The answer is $x^{-2}+y^{-2}=4c^{-2}$

The text you reported is not correct. It should be: "Find the locus of the CENTER OF THE circle passing through O, A and B". — Intelligenti pauca, Jan 29 '20 at 17:37
@Aretino I copied it exactly as it was. Even if it was wrong, I am still not getting the right answer — Aditya, Jan 29 '20 at 17:40
And the center is the midpoint of $AB$, but it is not true that $h^2+k^2=c^2$: just draw a sketch to understand why. — Intelligenti pauca, Jan 29 '20 at 17:41

Quanto · Answer 1 · 2020-01-29T20:56:07.047

Let C$(x,y)$ be the center of the line segment AB. Then, the coordinates of A and B are $(2x,0)$ and $(0,2y)$, respectively. Establish the relationship below with the area of the right triangle AOB,

$$Area_{AOB} - \frac 12 OA\cdot OB = \frac 12 c\cdot AB$$

or, in terms of the coordinate $x$ and $y$ for the center,

$$(2x)(2y) = c\sqrt{(2x)^2 + (2y)^2}$$

Rearrange to obtained the equation of the locus, which is shown in blue in the graph above

$$\frac1{x^2}+\frac1{y^2} = \frac4{c^2}$$

Find locus of center of circle passing origin, A and B, with AB at a constant distance $c$ from origin and intersecting axes at A and B.

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