# 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?

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

3. The Attempt at a Solution

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#### epenguin

Homework Helper
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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?

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

3. The Attempt at a Solution

View attachment 237530
[/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'.

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#### 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

Mentor
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

Homework Helper
Gold Member
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.

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#### Mark44

Mentor
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

Mentor
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

Homework Helper
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.

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#### epenguin

Homework Helper
Gold Member
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

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

#### SammyS

Staff Emeritus
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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

Gold Member
so can i get to know the solution?

#### OmCheeto

Gold Member
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.

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#### SammyS

Staff Emeritus
Homework Helper
Gold Member
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

Gold Member
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

Gold Member
Neilparker i can get some Tequila now...

#### ehild

Homework Helper
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

Homework Helper
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.

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#### chwala

Gold Member
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

Gold Member
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?

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#### chwala

Gold Member
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 ...

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#### chwala

Gold Member
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.

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