Calculate the Length of a Circle's Radius

[CroSinus]

1. Homework Statement
A circle in a x-y coordinate system is given. The equation of the circle is unknown. The circle touches the x axes at (3,0). It also passes through the point B (0,10). Calculate the length of the circle radius?

Homework Equations

(x-p)^2 + (y-q)^2 = r^2

The Attempt at a Solution

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[epenguin]

Homework Statement

A circle in a x-y coordinate system is given. The
equation of the circle is unknown. The circle touches the x axes at (3,0). It also passes through the point B (0,10). Calculate the length of the circle radius?

Homework Equations

(x-p)^2 + (y-q)^2 = r^2

The Attempt at a Solution

View attachment 237530 [/B][/QUOTE

The question says "the equation of the circle is unknown " but you quote and use it so I think you are not answering the question.

You will have to think geometry, and what you know of circles, and the things you can inscribe in them etc... tangents? Normals?

I don't know what your language is but learn once and for all how to spell 'Length'.

 

[phyzguy]

Using your equation you have written for the circle, the center of the circle is at the point (p,q). Look at your diagram. What can you say about p and q using the fact that the circle is tangent to the x-axis at the point (3,0)?

 

[Mark44]

The question says "the equation of the circle is unknown " but you quote and use it so I think you are not answering the question.

The equation of the specific circle of this problem is unknown. The equation given, $(x - p)^2 + (y - q)^2 = r^2$ is the general equation of a circle that is centered at (p, q) and with radius r. These three parameters are not given in this problem.

 

[epenguin]

I suppose you could read it like that. I took it to mean that you are not supposed to use the general equation of the circle, the wording does not seem to me to mean you just don't know the parameters. I would not take a risk on that interpretation in an exam.

Anyway it can be quite simply done without invoking this equation, just from lengths that are in your diagram; my previous suggestions were if anything overpowered.

 

[Mark44]

I took it to mean that you are not supposed to use the general equation of the circle

It seems pretty clear to me that "the equation of the circle is unknown." refers to the circle that is partially described in the first sentence below.

A circle in a x-y coordinate system is given. The equation of the circle is unknown.

I would be very surprised to find that the OP was not supposed to use the well-known general equation of a circle. If that were somehow the case, I would expect the problem to say explicitly, "Do not use the formula for the equation of a circle."

 

[YoungPhysicist]

Your question statement is a little bit odd.
If you are not supposed to use the general circle formula at all, then you'd have to prove it before you start.

But you are allowed to use the template, which is more likely, you can solve a simutaneous equation system from substituting the two known facts to two equations.

 

[Mark44]

Your question statement is a little bit odd.

The OP's native language is not English, so perhaps the translation into English is not quite equivalent to how the problem was originally stated.

 

[neilparker62]

Find equation of the perpendicular bisector of the chord drawn between the two given points. Then from your diagram the x co-ordinate of the centre is clear - just plug into equation to find y co-ordinate of centre which is also the radius as per your diagram.

Alternatively (and again referring to your diagram), simply apply Pythagoras theorem in the top triangle.

At a pinch you could even apply the cos rule to triangle ASB although this would require the use of a reduction formula - if you are familiar with those ? Generally easier to stick with right triangles as in second method above.

 

[epenguin]

Alternatively (and again referring to your diagram), simply apply Pythagoras theorem in the top triangle.

What I was refraining from saying. However OP not seen since Saturday evening, we shall see.

 

[chwala]

was this problem concluded? or its conclusion is dependant on op? i am interested in finding out the solution.

 

[SammyS]

was this problem concluded? or its conclusion is dependent on op? i am interested in finding out the solution.

Problem was not concluded, and OP apparently never responded - except for editing OP for the correct spelling of the word "length".

 

[chwala]

so can i get to know the solution?

 

[OmCheeto]

so can i get to know the solution?

I can tell you what the answer isn't.
When I did this problem in January, I used the numbers from the OP's doodle.
Unfortunately, they included an erroneous point: T2 (6,0)
I didn't catch the fact that it was a bogus number until I had finished the problem and arrived at the incorrect answer.
But I was able to determine my methodology from the order of the equations, and come up with the correct answer.
Unfortunately, we aren't allowed to simply give you the solution, so I'll provide one correct way to get the wrong answer:

I have fairly high confidence that this is the correct answer when the extra point, (6,0), is added to the problem.

It's basically the method recommended by neilparker62 in post #9.
The only thing I had to google was the fact that; A line perpendicular to another has a slope that is the negative reciprocal of the slope of the other line

I'm not a mathematician, so I don't know such things.

 

[SammyS]

so can i get to know the solution?

Of course you can. Give it a try.

I think a clue to the problem is the statement that the circle touches the x-axis at (3, 0). This is in contrast to what is stated regarding the circle passing through the point (0, 10).

 

[chwala]

Following post number 9 above- The equation of the perpendiculor bisector would be $y= 0.3x+4.55$
when $x=3, y=5.45$ therefore we have the points $(0,4.55)$ and $(3,5.45)$.The radius is given by $r^2= (3-0)^2 + (5.45-4.55)^2$ = $9.81$ it follows that $r=3.13$

 

[chwala]

Neilparker i can get some Tequila now...

 

[ehild]

Neilparker i can get some Tequila now...

Not yet. The radius you got is not correct. No need of any complicated derivation. Apply Pythagoras to the red triangle in the figure to calculate the distance OP which is equal to the radius r.

 

[neilparker62]

Apologies - not thinking! y-intercept of the perpendicular bisector is not a point on the circle. However since the circle is tangent to the x-axis, you can also use the tangent point (3,0) for your determination of the radius. Method given above is the most efficient since it solves for centre and radius at the same time.

 

[chwala]

Apologies - not thinking! y-intercept of the perpendicular bisector is not a point on the circle. However since the circle is tangent to the x-axis, you can also use the tangent point (3,0) for your determination of the radius. Method given above is the most efficient since it solves for centre and radius at the same time.

can you draw a full sketch of the circle with the triangle figure (inscribed) that's shown in post 19 using computer software, to enable me to have a rough idea of what i am trying to find?

 

[chwala]

In regards to your post 19.

Apologies - not thinking! y-intercept of the perpendicular bisector is not a point on the circle. However since the circle is tangent to the x-axis, you can also use the tangent point (3,0) for your determination of the radius. Method given above is the most efficient since it solves for centre and radius at the same time.

how is the circle tangent to the x- axis?

 

[ehild]

In regards to your post 19.

how is the circle tangent to the x- axis?

The problem text said

The circle touches the x axes at (3,0).

 

[chwala]

Attached find my free hand sketch of the circle... The roots seem to be negative.

The problem text said

I already responded to this with my Circle sketch, ...I can see Point (3,0) and the tangent ...

 

[chwala]

In regards to your post 19.

how is the circle tangent to the x- axis?

The circle is not a tangent to the x axis... But to the point (3,0)...that statement was a bit confusing.

 

[ehild]

The circle is not a tangent to the x axis... But to the point (3,0)...that statement was a bit confusing.

The circle touches the x-axis at point (3,0) does not mean it intersects the x-axis there. See https://www.math-only-math.com/circle-touches-x-axis.html and see my figure again

 

[chwala]

The circle touches the x-axis at point (3,0) does not mean it intersects the x-axis there. See https://www.math-only-math.com/circle-touches-x-axis.html and see my figure again

View attachment 246473

In your diagram the point (3,0) makes a tangent with the x axis. This is not correct. If that is the case, then the circle will not pass through points (0,10) as you have illustrated. The circle in my opinion will not form a tangent with the x axis, ...see my sketch.

 

[chwala]

The circle touches the x-axis at point (3,0) does not mean it intersects the x-axis there. See https://www.math-only-math.com/circle-touches-x-axis.html and see my figure again

View attachment 246473

There are only two possibilities, either the circles touches the point (3,0) implying that it's tangent to the point, which I respond by saying that's not possible or
Circle intersects the point (3,0) in order for it to pass through the point (0,10), which is highly likely as shown in my sketch...

 

[ehild]

In your diagram the point (3,0) makes a tangent with the x axis. This is not correct. If that is the case, then the circle will not pass through points (0,10) as you have illustrated. The circle in my opinion will not form a tangent with the x axis, ...see my sketch.

Try to read and understand the problem statement. And try to understand what a tangent line is. A point is not a tangent with a line as you said.
in your derivation, you just ignored l (the radius) and got imaginary roots for the position of the centre (r)
Replace l by r, and see what you get.

 

[ParticleGinger6]

Using general form (x-p)^2 + (y-q)^2 = r^2
if using a calculator you can find the equation that satisfies the following as (x-3)^2 + (y-5.45)^2 = 29.7025
sorry if this is already stated above

 

[chwala]

Try to read and understand the problem statement. And try to understand what a tangent line is. A point is not a tangent with a line as you said.
in your derivation, you just ignored l (the radius) and got imaginary roots for the position of the centre (r)
Replace l by r, and see what you get.

I don't understand how you use the points (3,r) and (0,10) and say that this is equal to r. What I have used is Pythagora's theorem where $adj^2+opp^2=hyp^2$and my hypotenuse is L. Unless am not seeing your view, enlighten me...

 

[chwala]

Let's assume for example that hypotenuse is $r=5$ then it follows that opposite side is $10-5=5$ and adjacent is $3$...doesn't make sense... it does not conform to Pythagoras theorem..

 

[ehild]

I quit

 

[SammyS]

In your diagram the point (3,0) makes a tangent with the x axis. This is not correct. If that is the case, then the circle will not pass through points (0,10) as you have illustrated. The circle in my opinion will not form a tangent with the x axis, ...see my sketch.

Here is the excellent figure given by my friend, @ehild .

The fact that the circle is tangent to the x-axis at (3, 0) is derived from the information in the problem statement indicating that the circle "touches" the x-axis at (3, 0) . This is in contrast to the wording for the other given point: "It also passes through the point B (0,10)."

I don't understand how you use the points (3,r) and (0,10) and say that this is equal to r. What I have used is Pythagora's theorem where $adj^2+opp^2=hyp^2$and my hypotenuse is L. Unless am not seeing your view, enlighten me...

Using @ehild's figure for the filled in right triangle we have:

One leg of the triangle has length of (10 − r). The length of the other leg is 3 units. Choose one of these as adj, the other as opp .

The hypotenuse, of course, has length of r .

You then find the solution by solving the following equation for r.

$(10 - r)^2 + 3^2 = r^2$

WolframAlpha gives the following graph when the resulting solution for r is plugged into the equation for a circle centered at (3, r) having a radius of length, r.

 

[chwala]

Here is the excellent figure given by my friend, @ehild .
View attachment 246494
The fact that the circle is tangent to the x-axis at (3, 0) is derived from the information in the problem statement indicating that the circle "touches" the x-axis at (3, 0) . This is in contrast to the wording for the other given point: "It also passes through the point B (0,10)." Using @ehild's figure for the filled in right triangle we have:

One leg of the triangle has length of (10 − r). The length of the other leg is 3 units. Choose one of these as adj, the other as opp .

The hypotenuse, of course, has length of r .

You then find the solution by solving the following equation for r.

$(10 - r)^2 + 3^2 = r^2$

WolframAlpha gives the following graph when the resulting solution for r is plugged into the equation for a circle centered at (3, r) having a radius of length, r.
View attachment 246493

So let's agree on this statement, the circle does not have a tangent with the $x-axis$ at point $(3,0)$

 

[chwala]

So let's agree on this statement, the circle does not have a tangent with the $x-axis$ at point $(3,0)$

I disagree with the fact that if point $(3,0)$ is tangent to x-axis that circle would pass through $(0,10)$

 

[SammyS]

I disagree with the fact that if point $(3,0)$ is tangent to x-axis that circle would pass through $(0,10)$

Have you solved the following equation for r ?

$(10 - r)^2 + 3^2 = r^2$

 

[chwala]

Have you solved the following equation for r ?

$(10 - r)^2 + 3^2 = r^2$

I don't understand... Enlighten me...

 

[chwala]

According to me show me how $(0-3)^2+(10-r)^2=r^2$...give for example $r=5$...the equation will not hold...

 

[SammyS]

Have you solved the following equation for r ?

$(10 - r)^2 + 3^2 = r^2$

I don't understand... Enlighten me...

Let me restate that.

Please, solve the following equation for $r$ .

$(10 - r)^2 + 3^2 = r^2$

It is derived from @ehild's figure, and the solution should give the radius of the circle in question, if said solution exists.

 

[SammyS]

According to me show me how $(0-3)^2+(10-r)^2=r^2$...give for example $r=5$...the equation will not hold...

It doesn't give r=5, so what's your point?

 

[chwala]

It doesn't give r=5, so what's your point?

Kindly, do you get my argument or rather confusion? I do not see how the points given satisfy the Pythagoras theorem...

 

[SammyS]

Kindly, do you get my argument or rather confusion? I do not see how the points given satisfy the Pythagoras theorem...

If you pick a radius of r = 5, and the circle is tangent to x-axis at (3, 0), then the circle passes through the point, (0, 9) .

... but, that's not this circle.

To pass through (0,10) and be tangent ... , radius will be greater than 5 .

 

[chwala]

If you pick a radius of r = 5, and the circle is tangent to x-axis at (3, 0), then the circle passes through the point, (0, 9) .

... but, that's not this circle.

To pass through (0,10) and be tangent ... , radius will be greater than 5 .

OK... Let me look at this tomorrow...

 

[SammyS]

Please solve for r.

The solution is not an integer, but it is rational.

 

[chwala]

Please solve for r.

The solution is not an integer, but it is rational.

Is $r$ a point on the Graph or is it a variable? That is an acronym for radius ' r ' of the circle?

 

[chwala]

Is $r$ a point on the Graph or is it a variable? That is an acronym for radius ' r ' of the circle?

 

[chwala]

Aaaaawwwaaah am so mad with myself. ehild I was blind I can see it now... You are right I can now find r. Bingo

 

[ehild]

r is the radius of the circle. You have to solve the equation $(10 - r)^2 + 3^2 = r^2$ to get it.

 

[chwala]

r is the radius of the circle. You have to solve the equation $(10 - r)^2 + 3^2 = r^2$ to get it.

True, I was supposed to think of distance between two points as opposed to the thinking of solving Pythagoras theorem, i got a bit confused here. Am OK with it now, will post solution later once I get on my desktop...

 

[chwala]

The circle touches the x-axis at point (3,0) does not mean it intersects the x-axis there. See https://www.math-only-math.com/circle-touches-x-axis.html and see my figure again

View attachment 246473

as per your approach the value of r= 5.45
i still have a problem with understanding your approach...sorry for bringing this discussion up, i believe its the whole essence of why the forum thrives. Consider the points (3,0) and (0,4), clearly the value of the hypotenuse here, presumably r = 5. Now going with your approach, let's have the same co ordinates, but now (3,r) and (0,4). What value will we get for r? are you implying that,
$3^2+(r-4)^2= r^2$? where
$r=3.125$?
make me understand your way of thoughts...