Linear algebra: Circles equation if you know two points on it.

Question

I am bashing my head to try to understand what this question is trying to show me about circles:

Show that for every $\lambda$ $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+\lambda[(x-x_1)(y_2-y_1)-(y-y_1)(x_2-x_1)]=0$$ is the equation for a circle through $P=(x_1,y_1), Q=(x_2,y_2)$.

The first part: $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$ Looks like the scalar product and equation for a circle, if P and Q were on opposite sides of the diameter.

The second part: $$\lambda[(x-x_1)(y_2-y_1)-(y-y_1)(x_2-x_1)]=0$$ is the equation for the line between P and Q.

How do I prove this and what is the intuition the problem wants to show me about circles and the points of intersection with a line?

I conjecture, this is just the equation for the statement "Angles in the same segment are equal": https://www.geogebra.org/m/vrgk9VYe — Konstantin, Oct 25 '21 at 13:38
I think you have a mistake in the first equation, in the square bracket. Both terms there contain $(y_2-y_1)$. I think one of them should be $(x_2-x_1)$ — Andrei, Oct 25 '21 at 14:33
If $P(x,y)$ is the current point and $A(x_1,y_1); B(x_2,y_2)$ are the points, the meaning of the statement is that the $PA\perp PB$, hence $AB$ is the diameter of the circle. — Vasile, Oct 25 '21 at 15:37

Andrei · Accepted Answer · 2021-10-25T15:49:53.320

1

Write your equations in terms of vectors, with $\vec u=(x-x_1,y-y_1)$ and $\vec v=(x-x_2,y-y_2)$. Then $\vec u-\vec v=(x_2-x_1,y_2-y_1)$. Then the first equation becomes $$\vec u\cdot \vec v+\lambda\vec k\cdot(\vec u\times(\vec u-\vec v))=0$$ We can write this as $$|\vec u||\vec v|(\cos\theta-\lambda\sin \theta)=0$$Here $\theta$ is the angle between vectors $\vec u$ and $\vec v$. Obviously the equation is satisfied when $\vec u=0$ or $\vec v=0$, so the curve passes through $(x_1,y_1)$ and $(x_2,y_2)$. The angular part being equal to zero just means that the tangent of $\theta$ is a constant: $$\tan\theta=\frac1\lambda$$ This means that the $\theta$ is a constant for every $\lambda$. Note that all the points in a circle see a chord at the same angle (and any point outside the circle see the chord at a different angle).

edited Oct 25 '21 at 15:49

answered Oct 25 '21 at 15:33

Andrei

37,370

$\sin{\theta}=k$ where k is a konstant gives us two possible circles. If we use $\tan{\theta}=k$ we have a unique circle since $\cos{\theta}$ is either negative or positive. I get that $\vec{u} \times \vec{v}$ is $\sin{\theta}$ and $\vec{u} \cdot \vec{v}$ is $\cos{\theta}$. But the angle between $\vec{u}$ and $\vec{v}$ is either $\theta$ or $180-\theta$ which gives two solutions to $\vec{u} \cdot \vec{v}$. – Erik Eriksson Oct 26 '21 at 11:30
I guess the fundamental problem I have with your explanation is that not every point on a circle sees the chord at the same angle. It is either $\theta$ or $180-\theta$ depending on wheter the center and point are on the same side of the line or not. – Erik Eriksson Oct 26 '21 at 12:35

score 0 · Answer 2 · answered Oct 25 '21 at 15:00

The curve clearly goes through both points $P$ and $Q$, (just plug in the coordinates for either point and the equation is satisfied).

The equation is of the form $$x^2+y^2 + \alpha(\lambda) x + \beta(\lambda) y + \gamma =0.$$

By completing the squares, you should be able to show that this can be written as

$$(x- a)^2 + (y-b)^2 = c^2,$$

which is a circle.

Linear algebra: Circles equation if you know two points on it.

2 Answers2