+490 Circles with centers $(2,1)$ and $(8,9)$ have radii $1$ and $9,$ respectively. The equation of a common external tangent to the circles can be written in the form $y=mx+b$ with $m < 0.$ What is $b?$
+2446 I think that this question is fairly difficult. I have added a few extra lines and points to the diagram where I saw fit:
Here are my initial observations in regard to this current diagram:
These initial observations are enough to get me started with this problem. I constructed a parallel line to the x-axis, \(\overleftrightarrow{AK}\). \(\angle BKA\) is, therefore, a right angle. Point K is located at (8,1) because it acquires the y-coordinate of Point A and the x-coordinate of Point B. \(\triangle BKA\) is a right triangle, and it is possible to find \(\tan(m\angle KAB)\).
\(\tan m\angle KAB=\frac{9-1}{8-2}=\frac{8}{6}=\frac{4}{3}\). Notice how 4/3 is also the slope of \(\overleftrightarrow{AB}\). The actual measure of the angle is not really important.
Because \(\overleftrightarrow{AK}||\overleftrightarrow{JH}\), \(\angle HJB\cong\angle KAB\). By extension, \(m\angle HJB=m\angle KAB\). This also means that \(\tan m\angle HJB=\tan m\angle KAB=\frac{4}{3}\).
Because two tangent lines of a circle intersect at an external point, J, \(\angle IJB\cong\angle HJB\). Using the same reasoning as before, \(\tan m\angle IJB=\tan m\angle HJB=\frac{4}{3}\). \(m\angle IJH=m\angle IJB+m\angle HJB=m\angle HJB+m\angle HJB=2m\angle HJB\)
Both tangent lines meet at an angle of \(2m\angle HJB\). Just like finding \(\tan m\angle KAB\) yielded the slope, \(\tan 2m\angle HJB\) will yield the slope of the desired line, \(\overleftrightarrow{IJ}\).
Luckily, we can utilize the Double-Angles Identities, specifically the \(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB}\).
| \(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB};\\ \tan m\angle HJB=\frac{4}{3}\) | This is the Double-Angle Identity for the tangent function. Replace all instances of \(\tan m\angle HJB\) with 4/3. |
| \(\tan 2m\angle HJB=\frac{2*\frac{4}{3}}{1-\left(\frac{4}{3}\right)^2}\\ \tan 2m\angle HJB=\frac{\frac{8}{3}}{\frac{-7}{9}}\\ \tan 2m\angle HJB=\frac{-24}{7} \) | Simplify. The result is the slope of the line \(\overleftrightarrow{IJ}\) . |
| \(y=\frac{-24}{7}x+b\) | This is the equation of the line \(\overleftrightarrow{IJ}\). |
The equation of the circle with the center A is \((x-2)^2+(y-1)^2=1\). Since \(y=\frac{-24}{7}x+b\), substitute that into the equation for the circle:
\((x-2)^2+(y-1)^2=1;\\ y=\frac{-24}{7}x+b \) | These are the equations we are grappling with. |
| \((x-2)^2+(\frac{-24}{7}x+[b-1])^2=1\) | It is time to expand this monstrosity. |
| \(x^2-4x+4+\left(\frac{-24}{7}x\right)^2+2*\frac{-24}{7}x*(b-1)+(b-1)^2=1\) | |
| \(x^2-\frac{28}{7}x+4+\frac{576}{49}x^2-\frac{48b-48}{7}x+b^2-2b+4=0\) | |
| \(\textcolor{red}{\frac{625}{49}}x^2\textcolor{blue}{-\left(\frac{48b-20}{7}\right)}x+\textcolor{green}{(b^2-2b+4)}=0\\ \textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}=0\) | Notice the parallelism between a quadratic and the current equation. Tangent lines always intersect a circle at only one point, so let's set the discriminant equal to zero to ensure only one solution is possible. |
| \(\Delta=b^2-4ac;\\ \Delta=\left(-\frac{48b-20}{7}\right)^2-4*\frac{625}{49}*(b^2-2b+4)\\ \Delta=\frac{-196b^2+3080b-9600}{49}\) | Set the discriminant equal to zero and solve. |
| \(\frac{-196b^2+3080b-9600}{49}=0\Rightarrow-196b^2+3080b-9600=0\) | Use the quadratic formula here. |
| \(b_1=\frac{30}{7}\text{ or }b_2=\frac{80}{7}\) | Reject b_2 because that is on the other side of the circle. |
+129852 I agree with X2.....this one is difficult.....!!!
I took an alternative approach....in the interest of space cosiderations, I will use some tecnology to help with the answer....but....I get the same answer as X2 did!!!
First of all..... I wanted to find the point where the tangent line intersects the x axis
Let this point = A
Call the distance between (2, 0) and A = x
So.. using similar triangles..we have this relationship
[ x ] / [ 6 + x] = 1 / 9
9x = 6 + x
8x = 6
x = 6/8 = 3/4
So....the point is ( 2 - x , 0) = ( 2 - 3/4, 0) = (1.25, 0) = A
Now we can find the distance betwwen the center of the smaller circle and A by using similar triangles again....call the distance, y....so we have
[ 1 + y] / [ 11 + y] = [ 1 ] / [ 9]
9 [ 1 + y ] = 11 + y
9 + 9y = 11 + y
8y = 2
y = 1/4 = .25
And we can find the distance between A and the point where the tangent line touches the smaller circle as sqrt [ (1 + 1/4)^2 - 1] = sqrt [ (5/4)^2 - 1] = sqrt [ (25 -16)/ 16] = sqrt (9/16) =
3/4 = .75
And we can find the point where the tangent line touches the smaller circle by using a little technology to find the intersection of these two circles :
(x - 2)^2 + ( y - 1)^2 = 1
(x - 1.25)^2 + y^2 = .75^2
(1.04, 0.72) = B
Likewise......we can find the distance between A and the point where the tangent line touches the larger circle as sqrt [ 11.25^2 - 9^2] = 6.75
And........We can find the point where the tangent line touches the larger circle by using a little technology to find the intersection of these two circles :
(x - 8)^2 + (y - 9)^2 = 81
(x - 1.25)^2 + y^2 = 6.75^2
( -0.64, 6.48) = C
The slope between B and C is
[ 6.48 - 0.72] / [ -0.64 - 1.04) = [ 5.76 ] / [-1.68] = -24/7
So....using A and this slope we can write an equation of the tangent line as
y = (-24/7)(x - 1.25)
y = (-24/7) (x - 5/4)
y = (-24/7)x + 120/ 28
y = (-24/7)x + 30/ 7
So...."b" = 30/7
+118678 That is a lot of work from both of you. Thanks Chris and XSquaredFactor.
That is some impressive LaTex too XSquared .
+33661 Here's my approach to this (I've also used some technology to solve the six simultaneous equations I obtained):
I should have noted that (x1, y1) are the coordinates of the point where the tangent line touches the larger circle, and (x2, y2) are the coordinates of the point where the tangent line touches the smaller circle.
.
