MHB How to Calculate the Radius of a Circle Tangent to Two Lines?

AI Thread Summary
To calculate the radius of a circle tangent to the lines y = 3x + 7 and y = 0.5x - 3, containing the point (8,1), the point of tangency is established at (8,1) for the line y = 0.5x - 3. The distance from the intersection of the tangent lines at (-4,-5) to (8,1) is approximately 13.416. Using the angle between the lines and the properties of right triangles, the radius can be determined as approximately 5.56. An algebraic approach also confirms this radius through the derivation of equations based on the circle's center and tangency conditions.
ginger
Messages
2
Reaction score
0
I'm having some difficulty with this problem and any help would be appreciated.

What is the radius of a circle tangent to the lines y = 3x + 7 and y = .5x - 3 and containing the point (8,1)?

I've determined that the given point (8,1) is the point of tangency of the line y = .5x - 3 and the circle. Also, the tangent lines intersect at (-4,-5) and the distance between (-4,-5) and (8,1) is approximately 13.416. It seems like I would need to find the center of the circle and use the Pythagorean Theorem to find the radius, but I can't figure out how to find the center. The solution is 5.56. Any suggestions?
 
Mathematics news on Phys.org
Re: tangent lines and circle

I would use the formula for the distance between a point and a line:

(1) $$d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}$$

which you can see derived here:

http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/

along with the formula for a circle in standard form:

(2) $$(x-h)^2+(y-k)^2=r^2$$

So, in (1) let $d=r$ and use the point $(h,k)$ to obtain:

$$r=\frac{\left|3h+7-k \right|}{\sqrt{3^2+1}}=\frac{\left|\frac{1}{2}h-3-k \right|}{\sqrt{\left(\frac{1}{2} \right)^2+1}}$$

and in (2) let $(x,y)=(8,1)$ to get:

$$(8-h)^2+(1-k)^2=r^2$$

From this, you have enough information to determine $r$.
 
Re: tangent lines and circle

I figured out another way to do it. I found the angle between the two lines using

tan(A - B) = (tan A - tan B)/(1 + tan A tan B)

So, using the slopes of the tangent lines, the angle between the lines is the inverse tangent of (3 - .5)/(1 +3(.5)), which gives an angle of 45 degrees. An angle bisector would go through the center of the circle, forming two right triangles, with the radius as one of the sides. Therefore, tan(22.5)=r/13.416, and r = 5.56.
 
Re: tangent lines and circle

ginger said:
I figured out another way to do it. I found the angle between the two lines using

tan(A - B) = (tan A - tan B)/(1 + tan A tan B)

So, using the slopes of the tangent lines, the angle between the lines is the inverse tangent of (3 - .5)/(1 +3(.5)), which gives an angle of 45 degrees. An angle bisector would go through the center of the circle, forming two right triangles, with the radius as one of the sides. Therefore, tan(22.5)=r/13.416, and r = 5.56.

I was on my way out when I posted my suggestion above, and so I did not actually try it, and it is cumbersome...the method you found is much more straightforward. Good job! (Sun)
 
We can find two circles satisfying the given requirements. You would have found the other radius by considering the other bisector. I will use an algebraic approach instead.

Given that the point (8,1) is on the circle, we know:

$$(8-h)^2+(1-k)^2=r^2$$

Since the circle is tangent to the line $$y=3x+7$$, we may write:

$$(x-h)^2+(3x+7-k)^2=r^2$$

$$(x-h)^2+(3x+7-k)^2=(8-h)^2+(1-k)^2$$

Expand and write in standard quadratic form

$$10x^2+(42-2h-6k)x+(16h-12k-16)=0$$

and require the discriminant to be zero:

$$(42-2h-6k)^2-4(10)(16h-12k-16)=0$$

$$(21-h-3k)^2-40(4h-3k-4)=0$$

$$h^2+6hk-202h+9k^2-6k+601=0$$

Doing the same with the other line, we eventually find:

$$k=17-2h$$

Now, substituting this into the first equation, we find:

$$h^2-28h+124=0$$

$$h=14\pm6\sqrt{2}$$

Case 1: $$h=14+6\sqrt{2}$$

$$k=-\left(11+12\sqrt{2} \right)$$

$$r=6\left(\sqrt{5}+\sqrt{10} \right)\approx32.390073826009015$$

Plot of lines and resulting circle:

View attachment 1031

Case 2: $$h=14-6\sqrt{2}$$

$$k=12\sqrt{2}-11$$

$$r=6\left(\sqrt{10}-\sqrt{5} \right)\approx5.55725809601154$$

Plot of lines and resulting circle:

View attachment 1032
 

Attachments

  • ginger1.jpg
    ginger1.jpg
    8.5 KB · Views: 107
  • ginger2.jpg
    ginger2.jpg
    9.2 KB · Views: 94
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
2
Views
2K
Replies
4
Views
1K
Replies
2
Views
2K
Replies
10
Views
3K
Replies
4
Views
2K
Back
Top