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There is an equation which instantly gives you the chord having mid point $(x_1,y_1)$ in a circle discussed here: For a circle,

$$ C(x,y)=x^2 +y^2 + 2gx + 2fy + c = 0$$

The chord having mid point $(a,b)$ is given as:

$$ ax+by+g(x+a)+f(y+b) + c = C(a,b)$$

and I found a similar result here. However is there a general elegant way to prove this equation for all conics?

Bernard
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tryst with freedom
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2 Answers2

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Consider the equation of a general conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$

Now consider a point $(m,n)$ which is to be taken as the midpoint of the chord of the conic.

Now consider the equation $$\begin{aligned}amx+h(my+nx)+bny+g(x+m)+f(y+n)+c \\ =am^2+2hmn+bn^2+2gm+2fn+c\end{aligned}$$

Above is a straight line which passes through $(m,n)$ and of course it would cut the conic at maximum of $2$ points. Now if we solve this equation with the conic, and consider that as a quadratic in $x$, we will get the sum of roots as $2m$ and similarly if we consider it a quadratic in $y$, sum of roots would be $2n$.

Edit: Note that if the roots of the quadratic are real and distinct, then a chord exists, if the roots are equal, then the point $(m,n)$ lies on the conic and hence that line would be a tangent. If the roots are imaginary, no chord exists.

V.G
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    How do you know for sure it cuts conic at two points? the later calculations you've stated to do are complex – tryst with freedom Feb 06 '21 at 11:18
  • You don't have to do any calculations for showing that there would be maximum of $2$ points because it is a quadratic and so maximum 2 roots, but even leave that, imagine any conic you can, can you think of any where a line cuts it at 3 points? It is because a conic is a 2 degree curve. Conics are: Parabola, Hyperbola, Circle and Ellipse. – V.G Feb 06 '21 at 11:19
  • You need to check the discriminant to say that for sure @Light Yagami – tryst with freedom Feb 06 '21 at 11:29
  • I need to check the determinant of what and to say what? Please write full sentence... – V.G Feb 06 '21 at 11:30
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    That if we solve that equation with conic that we will get two roots, you would have to check discriminant to see if there exists two roots, which is tedious calcs – tryst with freedom Feb 06 '21 at 11:34
  • Well, it is not at all necessary to have two real roots since the point $(m,n)$ is general. Of course, if the point lies outside the conic, we cannot have any chord in that case, but why do we need to check the roots? A quadratic means two roots maximum, real or imaginary, it doesn't matter if they are imaginary. Important thing one should focus on here is their sum, which would be $2m$ if considered a quadratic in $x$ and similarly for $y$, Implying that $(m,n)$ is the midpoint of the chord. – V.G Feb 06 '21 at 11:37
  • Suppose you didn't know this was equation of chord and I gave you this equation for first time in your life, then you would not be able to say if it passes through conic two times. Only way to check is by plotting/ checking discriminant – tryst with freedom Feb 06 '21 at 12:34
  • I don't understand why you are not getting the point. It is not about me knowing the equation, I am using the fact that any line will cut a conic section at maximum of $2$ points. And we don't need to check any graph or discriminant! – V.G Feb 06 '21 at 12:41
  • counter example: Consider a line $y=10$ and the unit circle $x^2 +y^2 =1 $ – tryst with freedom Feb 06 '21 at 12:42
  • Yeah, so? I said maximum, not exact. What is counter in it? – V.G Feb 06 '21 at 12:43
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    My point is that without checking the discriminant we can't guarantee it's a chord, if it's maximum is two points, it doesn't tell us if t's chord or not. – tryst with freedom Feb 06 '21 at 12:44
  • And my point is that it would be a chord if the point $(m,n)$ lies inside the conic. If the point is not inside, we cannot have any chord. What are you arguing about? You asked for an equation of a chord having the mid point $(m,n)$, it is only possible if it exists. Otherwise no equation exists. – V.G Feb 06 '21 at 12:47
  • If you want to ensure that the chord exists it is then you will need the discriminant and you would get conditions on $m$ and $n$. – V.G Feb 06 '21 at 12:48
  • @Buraian I even thought of a one line step to see if the chord exists or not, we don't need discriminant. Just put the point in the conic and see it's sign, from there you will come to know whether the point lies inside or outside. For example, we have the circle $x^2+y^2=1$, If the given point is $1,2$, then we have $1^2+2^2-1>0$ hence it is outside the conic, for any point inside it, it should come negative. Similarly, if it is a parabola say $y^2=4x$, the point $(2,2)$ lies inside it because $2^2-4\cdot 2<0$. And hence, if $$am^2+2hmn+bn^2+2gm+2fn+c<0$$ point is inside & hence chord exists. – V.G Feb 06 '21 at 13:04
  • Let us continue this discussion in chat. – tryst with freedom Feb 06 '21 at 14:20
  • @LightYagami is there any intuition as to why you want us to consider that specific equation? – DatBoi Feb 06 '21 at 16:11
  • @LightYagami oh I see. How exactly do we use derivatives to obtain it? A bit of an insight would be great. – DatBoi Feb 06 '21 at 16:33
  • @DatBoi: I am sorry, I was thinking about writing equation of tangent. No derivatives here. You can read about pole and polar. That is from where this comes. – V.G Feb 06 '21 at 16:42
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Actually there is a very easy way to do this which I realized when I saw the question again.

I take the equation of circle as $C(x,y) = x^2 + y^2 + 2gx + 2fy +c$

The center of circle is $(-g,-f)$, we can write the normal vector to the line segment made by chord as:

$$ n = (x_m + g , y_m +f)$$

That is the vector connecting center of circle to mid point of chord. We can also write a vector along the chord as:

$$ r = (x-x_m,y-y_m)$$

We have:

$$ n \cdot r = 0 \iff (x-x_m) (x_m+g) + (y_m +f)(y-y_m) =0 \iff xx_m + xg -x_m^2 - x_m g + y_m y -y_m^2 +fy -fy_m=0$$

Rearranging terms in the double implication, we have:

$$ xx_m + yy_m +gx + fy = x_m^2 + y_m^2 + fy_m + gx_m$$

Adding $c$ on both sides, we can write:

$$ xx_m + y y_m + gx +fy + c = C(x_m,y_m)$$

Which is what was required.

tryst with freedom
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