Question

Circles with centers (2,4) and (14,9) have radii 4 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?

Answer 1

here is a picture model ^

Answer 2

Calculus?

Answer 3

nope

Answer 4

its sophomore math team

Answer 5

shouldnt involves anything harder than trig

Answer 6

I think that b is 8. Isn't that what you need to know?

Answer 7

yes but it is not 8 it is a fraction close to it

Answer 8

i have deduced that the slope intercept form of the equation of the line connecting the centers of the circles is y=5x/12 + 19/6

Answer 9

but im stuck after that

Answer 10

Label the points of tangency (x1,y1) and (x2,y2). The slopes of the radius through those points are equal. Further note that the slope of the tangent line is the opposite of the reciprocal of that slope. Manipulate these three facts a little, and you should be able to set up a system of linear equations that is solvable. It is a little ambitious, but you should be able to do it.

Answer 11

perpendicualr lines have negative reciprocal slopes correct?

Answer 12

Yep.

Answer 13

i cant seem to get them into a form where they can be compared. do you have any hints as to what i need to solve for? should i start with slope?

Answer 14

After looking at it for a little while, I have five equations and five variables. The solution doesn't look very simple, but seems possible. Equations follow:\[\frac{y_1-4}{x_1-2}=\frac{y_2-9}{x_2-14}=m\]\[\frac{y_2-y_1}{x_2-x_1}-\frac{-1}{m}\]\[(x_1-2)^2+(y_1-4)^2=16\]\[(x_2-14)^2+(y_2-9)^2=81\]To solve this doesn't seem like a straightforward affair; you will have to rearrange these equations and substitute repeatedly. But it seems possible.

Answer 15

i have been at this for a while and solving is incredibly difficult

Answer 16

also how exactly did you deduce the final two equations. they look like circles but?

Answer 17

Note that the first line has at least two equations. By the way, I'm not at all sure this is the right way to solve this; the calculus solution would seem to be simpler.

Answer 18

what is the calculus solution? what does it involve as far as mathematical concepts? would i understand?

Answer 19

The points (x1,y1) and (x2,y2) are particular points on the circle, so they meet the specification of the circle.

Answer 20

calculus solution involves integrals etc? cuz that stuff i am not familiar with but am trying to learn on the side (with much difficulty because i have not yet had alg 2 or precalc)

Answer 21

I doubt I would be able to explain this over the internet. Suffice to say, we would find functions for the top half of both circles, then differentiate to find the slope as a function of x. From there, we set the slopes equal and use a couple of the equations above to link the whole mess together. As I look at it further, it doesn't seem too friendly, either.

Answer 22

let's say i found the degree measure of the line of external tangency with respect to the x-axis (the acute angle formed by the line of tangency an the x-axis) would this help to find the slope? these equations are killing me and im certain there is a way around

Answer 23

That would give you the slope. That would reduce the problem somewhat, but you will still have quite a few variables to solve for before you can get the answer.

Answer 24

ok well here is my theory: the angle formed by the line connecting the two centers is equal to z. tan(z) equals 5/12, so arctan(5/12) is the measure of the angle formed by the line connecting the centers to the x axis. Becasue this line is connecting centers, and the two lines of external tangency to the circle are: the x-axis, and the undefined line, the line connecting the centers should BISECT the angle formed by the connection of the external line of tangency and the x-axis (the line of centers bisects the angle formed by the intersection of the two external tangents). AS a resULT, tan (2z) is the angle measure of the line of centers and the external tangent. Is this methodology correct and how can i deduce the slope from this? I also believe that tan2(z) is equal to a fraction with 2 tan (z) as the denominator but ???????

Answer 25

\[m = \tan 2z = \frac{ 2 \tan z }{ ? }\]

Answer 26

i cant use a calculator soooooo i need this to be a manipulatable fraction

Answer 27

just saying that with the solution to this fraction, the problem cracks wide open and i solve in 2 steps

Answer 28

Not sure. I've been looking at a slightly different (geometry) approach to the problem.

Connect the centers of the circles and extend the line.

Note that the line of tangency and the line through the centers intercept off to the left (second or third quadrant).

Make similar triangles to find out how far from the center of the smaller circle the point of intersection of these two lines is (I'll call it z). I think it comes out like\[\frac{x}{4}=\frac{13+z}{9} \implies z=\frac{52}{5}\]

Now note that z is the hypotenuse of a right triangle, and the angle between the two lines (call it theta) is \[\sin \theta = \frac{4}{\frac{52}{5}}=\frac{5}{13} \implies \theta = \sin^{-1} \frac{5}{13}\]

Add theta to the angle of the line through the centers, and you have the slope.

Answer 29

Note that I called typed x for z at the beginning of the first equation.

Answer 30

what do you think the slope is? all i need is the slope of this external tangent, and the problem is solved

Answer 31

i suppose it is solvable with a calculator but that isnt allowed

Answer 32

Sorry, I haven't been too much help here. This is an interesting problem. Hope you can find a way to get it figured out. I've gotta go now.

Answer 33

alright see ya

Answer 34

both the line of centers and the external tangent go through the point \[\frac{ -38 }{ 5 }, 0\]

Answer 35

i just need the dammmmm slope!!!!!!!!!!!!

Answer 36

You've got it. The angle between the tangent line and the x axis is twice twice the angle between the line through the centers and the x axis. Find the slope of that line and double it. There is plenty of information to get that.

Answer 37

Really have to go. Good luck. Do math every day.