Calculate the Length of a Circle's Radius

[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...

 

[SammyS]

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...

I will respond for @ehild.
(Normally, I would be very reluctant to speak for another Helper. - It is perhaps 5:30 am in Hungary as I post this.)

Let's suppose that the question is to find the (length of the) radius of a circle which is tangent to the x-axis at (3,0) and the point (0,4) is also on the circle.

Fantastic! You have found the length of the radius of such a circle to be 3.125 units.

Let's consider some details.
The center of the circle is at (3, 3.125).

The distance of the center from (3, 0) is 3.125.

What is the distance from the center to the point (0,4) ?

 

[chwala]

I can see that we get the same radius of 3.125. Thanks Sammy, bingo!

 

[ehild]

I will respond for @ehild.
(Normally, I would be very reluctant to speak for another Helper. - It is perhaps 5:30 am in Hungary as I post this.)

Thank you Sammy for speaking for me. I just awoke that time.

 

[aheight]

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.

I dont' see how the law of cosines can be used to find the radius of the circle with the information given. I assume we have it tangent at (3,0) and the reference is the triangle in the picture below. Might you or someone else explain please?

 

[SammyS]

I dont' see how the law of cosines can be used to find the radius of the circle with the information given. I assume we have it tangent at (3,0) and the reference is the triangle in the picture below. Might you or someone else explain please?
View attachment 248112

It works out in a rather straight forward manner (with the Law of Cosines).

Have you tried to do this with the Law of Cosines?
If not, then try it.
If you have, then show us your attempt.

 

[aheight]

It works out in a rather straight forward manner (with the Law of Cosines).

Have you tried to do this with the Law of Cosines?
If not, then try it.
If you have, then show us your attempt.

Ok.

Using the figure below I have using the Law of Cosine:
$$
\begin{aligned}
a^2&=b^2+c^2-2bc\cos(\theta)\\
a^2&=2r^2-2r^2\cos{\theta} \\
a^2&=2r^2(1-\cos{\theta}) \\
\end{aligned}
$$
and we know $a=\sqrt{109}$ so we have $1/2(109)=r^2(1-\cos{\theta})$. And using the hint that a reduction formula would be needed, I can use $\frac{1-\cos{\theta}}{2}=\sin^2{\theta/2}$ so that

$$(109)=4r^2\sin^2{(\theta/2)}$$
so
$$ r=\frac{\sqrt{109}}{2 \sin{(\theta/2)}}$$

and I can find $a/2$ with the mid-point formula. However, I would need to know the center in order to compute the length of the red line segment so that I could then figure out what $\theta$ is. And in order to find the center I would have needed to solve for the radius using the first version of the solution stated in the reference using $$(10-r)^2+9=r^2$$.

I guess I was asking how do I use the cosine law without first using the two methods before this method?

 

[SammyS]

I guess I was asking how do I use the cosine law without first using the two methods before this method?

I think you will find that it is more direct to use the angle formed by side $a$ and side $c$.

The cosine of that angle can be found quite easily and is $\dfrac{10}{\sqrt{109}}$ .

Calling this angle, $\phi$, and applying the Law of Cosines gives:

$b^2 = a^2 + c^2 - 2ac\cos(\phi)$, where $b=c=r$ and $a=\sqrt{109}$.

 

[aheight]

Ok thanks for that. It is simple your way. I did it using the diagram below using your help with $\cos(\beta)$:

$\sin(\alpha)=10/\sqrt{109}$ and $\beta=\pi/2-\alpha$ and $\cos(\pi/2-\alpha)=\sin(\alpha)$ and $\cos(\pi/2-\alpha)=\sin(\alpha)$ and therefore
$\theta/2=\pi/2-\arccos(10/\sqrt(109)\approx 1.279$ so then using my expression above, I get:

$$
r=\frac{\sqrt{109}}{2\sin(1.279)}\approx 5.45
$$

which is consistent with the results above.

Thanks Sammy. This teaches me to first write down all the information I can compute first before trying to solve the problem. Interesting problem.

 

[WWGD]

But it gets kind of tricky possibly. Can we assume it touches the x-axis _only_ at (3,0) so that (3,0) is on the circle? It seems this would give the solution.Edit:That did not come out right. I meant the circle may interest t the x-axis more than on e unless it is explicitly stated otherwise.

 

[256bits]

But it gets kind of tricky possibly. Can we assume it touches the x-axis _only_ at (3,0) so that (3,0) is on the circle? It seems this would give the solution.Edit:That did not come out right. I meant the circle may interest t the x-axis more than on e unless it is explicitly stated otherwise.

An infinite number of circles then, the smallest of which would have the diameter of distance between the 2 points given.

 

[neilparker62]

Ok.

Using the figure below I have using the Law of Cosine:
$$
\begin{aligned}
a^2&=b^2+c^2-2bc\cos(\theta)\\
a^2&=2r^2-2r^2\cos{\theta} \\
a^2&=2r^2(1-\cos{\theta}) \\
\end{aligned}
$$

cos(θ) = cos(90 + δ) = -sin(δ) where sin(δ) = (10-r) / r.

or produce the radius forming a diameter 2r. Then:

a/2r = cos(β)=sin(α)=10/a

 

[kuruman]

I took upon myself to summarize a solution for the benefit of anyone you might look at this thread and try to sift through the various approaches for getting to the answer. I think the most straightforward strategy has been outlined in post #14.
1. Find the equation of the line connecting points A {0, 10} and B {3,0} that lie on the circle. Can be done by inspection.
Answer: $y=-\frac{10}{3}x+10.$
2. Find the equation of the perpendicular bisector of segment AB. The slope is the negative inverse of the slope of AB, $m=\frac{3}{10}$. It must pass through the midpoint of AB, point C {$\dfrac{3}{2},5$}. To find the intercept $b$, solve $$5=\frac{3}{10}\times \frac{3}{2}+b ~\rightarrow~b=\frac{91}{20}.$$Answer: $y=\dfrac{3}{10}x+\dfrac{91}{20}.$
3. Find the intersection of the perpendicular bisector and the line $x=3$. Just plug in $x=3$ in the equation found in step 2.
$$y=\frac{3}{10}\times 3+\frac{91}{20}=\frac{109}{20}=5.45.$$Answer: The center of the circle is at point O {3, 5.45} and the radius of the circle is $R=5.45.$

 

[chwala]

Thanks Kuruman, I will look at it in depth later...