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Tangent lines and circle

ginger

New member
Jul 19, 2013
2
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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: tangent lines and circle

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

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

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

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

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

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

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

From this, you have enough information to determine $r$.
 

ginger

New member
Jul 19, 2013
2
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.
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)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

Since the circle is tangent to the line \(\displaystyle y=3x+7\), we may write:

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

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

Expand and write in standard quadratic form

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

and require the discriminant to be zero:

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

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

\(\displaystyle h^2+6hk-202h+9k^2-6k+601=0\)

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

\(\displaystyle k=17-2h\)

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

\(\displaystyle h^2-28h+124=0\)

\(\displaystyle h=14\pm6\sqrt{2}\)

Case 1: \(\displaystyle h=14+6\sqrt{2}\)

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

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

Plot of lines and resulting circle:

ginger1.jpg

Case 2: \(\displaystyle h=14-6\sqrt{2}\)

\(\displaystyle k=12\sqrt{2}-11\)

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

Plot of lines and resulting circle:

ginger2.jpg