Explanation needed circle equation

A

Ana.stasia

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So I have a point A(x, 0) I would like to know why that makes k=n=0.

p, q are the cordinates of the center of the circle.
k, n are from this form of an equation: y=kx+n

IMG_20210313_232108.jpg
 
I do not understand your question at all.

Try giving the original problem completely and exactly.
 
Determine the parameter k such that the circle x2 + y2- (5k-1) x + (4-2k) y = 5k touches the x-asis

The solution I was given
Screenshot_20210314-003503.jpg

The part I described is in the top right corner under the graph.
 
Last edited:
I don't see what that part has to do with the rest of the work. It appears to be showing that [MATH]q = \pm r[/MATH], which to me is obvious as the center has to be directly above the point of tangency; and that feeds into the rest of the work. But I can't see where

1615684320591.png

comes from, or where there is any mention of the line y = kx + n that you mentioned, or how k=n=0 comes from that, or what that has to do with a problem in which we find that h is -1/5, not 0.

Surely there must have been some sort of spoken explanation of this that was not caught in the writing. Or is it left over from a different problem?
 
Dr.Peterson said:
I don't see what that part has to do with the rest of the work. It appears to be showing that [MATH]q = \pm r[/MATH], which to me is obvious as the center has to be directly above the point of tangency; and that feeds into the rest of the work. But I can't see where

View attachment 25707

comes from, or where there is any mention of the line y = kx + n that you mentioned, or how k=n=0 comes from that, or what that has to do with a problem in which we find that h is -1/5, not 0.

Surely there must have been some sort of spoken explanation of this that was not caught in the writing. Or is it left over from a different problem?
Click to expand...

I am not really sure myself. That's the formula that is correct if a is tangent to the circle.
 
I guess the question is:

The circle with equation: [MATH]x^2+y^2-(5k-1)x+(4-2k)y=5k[/MATH] touches the [MATH]x[/MATH]-axis, (e.g.) as shown:
1615722783454.png (i.e. the [MATH]x[/MATH]-axis is a tangent to the circle).
Find [MATH]k[/MATH].

The work then shows us that this is a circle with centre [MATH](p,q)[/MATH] and radius [MATH]r[/MATH], with:
[MATH]p=\frac{5k-1}{2},\ \ q=k-2,\ \ r^2=5k+\left(2-k\right)^2+\ \left(\frac{5k-1}{2}\right)^2[/MATH]

Now the little bit under the picture.
It is a fact that when [MATH]y=Kx+n[/MATH] is a tangent to a circle with centre [MATH](p,q)[/MATH] and radius [MATH]r[/MATH], then
[MATH]\boldsymbol{r^2\left(1+K^2\right)=\left(Kp-q+n\right)^2}[/MATH](lamentably a different [MATH]K[/MATH] from the [MATH]k[/MATH] used in the rest of the question)!
(It is an interesting question to prove this formula).

So in this case, since the equation of the tangent line (the [MATH]x[/MATH]-axis) is [MATH]y=0[/MATH], then [MATH]K=0[/MATH] and [MATH]n=0[/MATH].
Therefore the formula [MATH]\boldsymbol{r^2\left(1+K^2\right)=\left(Kp-q+n\right)^2}[/MATH] reduces to [MATH]\boldsymbol{r^2=q^2}[/MATH].

However as Dr.Peterson points out, it is obvious in this case that [MATH]r=|q|[/MATH] (and therefore [MATH]r^2=q^2[/MATH]) since the centre of the circle is at a height [MATH]|q|[/MATH] above or below the [MATH]x[/MATH]-axis and this is the radius of the circle:
1615722798727.png

Using [MATH]r^2=q^2[/MATH], the work gives [MATH]k[/MATH]=-1/5
 
lex said:
It is a fact that when [MATH]y=Kx+n[/MATH] is a tangent to a circle with centre [MATH](p,q)[/MATH] and radius [MATH]r[/MATH], then
[MATH]\boldsymbol{r^2\left(1+K^2\right)=\left(Kp-q+n\right)^2}[/MATH](lamentably a different [MATH]K[/MATH] from the [MATH]k[/MATH] used in the rest of the question)!
(It is an interesting question to prove this formula).
Click to expand...
Are you saying that this is a formula students are taught to memorize? Wow.
 
Dr.Peterson said:
Are you saying that this is a formula students are taught to memorize? Wow.
Click to expand...
I must say I had never heard of it. I just read it on Ana.stasia's post and then proved it. By it's use in this question it would appear to be something they are supposed to have known. In any case, it seems to be the explanation for the enigmatic scribbles under the diagram.
 
Of course, this question could have been done in a few lines, using the method used in your other thread:
Parameter k, circle equation
Substitute ([MATH]x[/MATH],0) into [MATH]x^2+y^2-\left(5k-1\right)x+\left(4-2k\right)y-5k=0[/MATH] and find when it has repeated roots.
I.e. when does [MATH]x^2-(5k-1)x-5k=0[/MATH] have repeated roots?
Solve [MATH]\left(5k-1\right)^2=4\left(-5k\right)[/MATH][MATH]25k^2+10k+1=0[/MATH][MATH]\left(5k+1\right)^2=0[/MATH][MATH]k=-\frac{1}{5}[/MATH]
 
lex said:
Of course, this question could have been done in a few lines, using the method used in your other thread:
Parameter k, circle equation
Substitute ([MATH]x[/MATH],0) into [MATH]x^2+y^2-\left(5k-1\right)x+\left(4-2k\right)y-5k=0[/MATH] and find when it has repeated roots.
I.e. when does [MATH]x^2-(5k-1)x-5k=0[/MATH] have repeated roots?
Solve [MATH]\left(5k-1\right)^2=4\left(-5k\right)[/MATH][MATH]25k^2+10k+1=0[/MATH][MATH]\left(5k+1\right)^2=0[/MATH][MATH]k=-\frac{1}{5}[/MATH]
Click to expand...

Thank you. I understand now.
 
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