Geometrical interpretation of a power of a point

[Mondo]

Hello, I try to get the algebraic proof of this theorem. I follow this https://i2.paste.pics/796e202e27015761ebe45897e885b151.png. I understand how the author derived the final solution for the case of two solutions, namely PQ and PR but I don't understand the case of a single solution. How can that be [MATH]PT^2[/MATH] when the solution of a quadratic equation for a single root is just [MATH]\frac{-b}{2a}[/MATH] which for this case is just [MATH]-1[/MATH].
Also, since we assume [MATH]C(x,y) = 0 [/MATH] does it mean the product of root (two root) or a square of a single root is equal to 0?

Thanks

 

[lex]

C(x,y) is not assumed to be 0. It is [MATH] C(x^{'},y^{'}) = 0 [/MATH]


Why do you think [MATH]\tfrac{-b}{2a}=-1[/MATH]?
The quadratic equation in [MATH] r [/MATH] is:

 

[Mondo]

Why do you think −b2a=−1−b2a=−1\displaystyle \tfrac{-b}{2a}=-1?

I got confused by the figure 18.1 which clearly shows one root - the tangent has one point of contact with the circle. However the discriminant of the quadratic equation is non zero hence we must have two root. Now, since we know there is only one we have a situation where we have a double root - two same roots. Do I understand that correctly now?

Hence, vieta theorem still holds and we have [MATH]r_1 * r_2 = C(x,y) = PT^2[/MATH]

 

[lex]

The question is framed so that the line does intersect the circle, so there are either two distinct roots or there is one 'double' root (the two roots are the same). The discriminant of the quadratic equation in [MATH] r[/MATH]:
[MATH]r^{2}+2r(x\cos \theta + y \sin \theta + g \cos \theta + f \sin \theta) + C(x,y)=0 \hspace2ex[/MATH] would then be >0 or 0, respectively.

In the case of two distinct roots, the discriminant is >0
Using the Vieta formulas, the product of the distinct roots: [MATH]PQ.PR[/MATH] is the constant term [MATH]C(x,y)[/MATH]([MATH]r_1=PQ \text{, }r_2=PR[/MATH], say).

In the case of a tangent, there is one, 'double' root (the 'two roots' are the same) and the discriminant is 0.
Using the Vieta formulas again, the product of the double root: [MATH]PT^{2}[/MATH] is the constant term [MATH]C(x,y)[/MATH]([MATH]r_1=r_2=PT).[/MATH]

 

[Mondo]

@lex, all is clear beside this:

In the case of a tangent, there is one, 'double' root (the 'two roots' are the same) and the discriminant is 0.

How do you know there is one, 'double root'? Or maybe first, how do you know the discriminant is 0? Do you get that by algebraic calculation or from visual inspection of a Figure 18.1? I bet the later since in algebraic calculation we have an unknown term [MATH]C(x,y)[/MATH]

 

[lex]

We know from the nature of a tangent to a circle that it only intersects the circle once, at the point of tangency. (Fairly easy to convince yourself of this and/or prove it). Therefore we know that the quadratic equation in r, giving the intersection points has one and only one root ('double' root). Therefore we know that the discriminant must be 0.

 

[Mondo]

lex, thanks for explanation. I got confused just because of the fact that tangent only touches the circle in one point -- hence I assumed one root and that;s why I was unable to apply vieta formula. On the other hand if I had a clear equation like (x-a)(x-a) = 0 then I would without any problem come up with double root solution. So now, how can we make ourself sure, having only the diagram that there are two root even though we see one.

Thanks

 

[lex]

how can we make ourself sure, having only the diagram that there are two roots even though we see one.

We don't solve this with only the diagram. The word 'root' implies that we have an equation. The 'double root' is a feature of the equation.
For the tangent case - for the diagram, it is clear there is one solution, - for the quadratic equation, this means there is a 'double root'.

In fact, the question can be done without the diagram.
If you notice, in the working they say "and if [MATH](x' \text{, } y'[/MATH]) is the point Q." (Not R or 'Q and R').
So we are saying that - if Q is a point of intersection and r is the distance PQ, etc... then:
[MATH]r^2+2r(x \cos \theta + y \sin \theta + g \cos \theta + f \sin \theta) + C(x, y)=0[/MATH]It is our knowledge of quadratic equations which allows us to say that this has potentially 2 roots, a double root (one solution) or no roots.
We could ask for a geometrical interpretation of these 3 situations (in the context of this particular question). Then, with reference to a diagram, we could say the scenarios are:

(By the way, I do not recommend doing a question without a 'picture'. I'm just trying to indicate that the 'double root' idea comes from the equation. A picture is an important tool in solving a problem).

 

[Mondo]

@lex, well explained. Thanks a lot!

 

[lex]

Once I re-read it, it doesn't look so clear, but hopefully it helped in any case.