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Take the intersection of line AB and CD as point E.

EB = ED as tangents from a single point to a circle are equal

EB = EC + CD take EC = x
=> EB = x+ 24

Also AE=EC = x using same explanation above.

AB = AE+EB = x+x+24 = 2x+24 = even number- if you are short of time you can easily guess.

Now triangle ECO is similar to triangle triangle EDO' where O and O' are center of 1st and 2nd respectively.

=>\(\frac{EC}{OD} = \frac{O'D}{ED}\)

=> \(\frac{x}{5}= \frac{5}{(x+24)}\)

=> \(x^2 +24x -25 = 0\)
=> \((x +25)*(x-1)= 0\) => \(x =1\)

=> AB\(= 2x*+24 = 2+24 = 26\)
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gurpreetsingh
Take the intersection of line AB and CD as point E.

EB = ED as tangents from a single point to a circle are equal

EB = EC + CD take EC = x
=> EB = x+ 24

Also AE=EC = x using same explanation above.

AB = AE+EB = x+x+24 = 2x+24 = even number- if you are short of time you can easily guess.

Now triangle ECO is similar to triangle triangle EDO' where O and O' are center of 1st and 2nd respectively.

=>\(\frac{EC}{OD} = \frac{O'D}{ED}\)

=> \(\frac{x}{5}= \frac{5}{(x+24)}\)

=> \(x^2 +24x -25 = 0\)
=> \((x +25)*(x-1)= 0\) => \(x =1\)

=> AB\(= 2x*+24 = 2+24 = 26\)


the only question I have is how did you arrive at Triangle Similarity?

we have 1 Smaller Right Triangle OCE ---- Leg = x ; Leg = 5

and 2 Larger Right Triangle EDO' ----- Leg = x + 24 ; Leg = 5

How do you find that these 2 Triangles are Similar?

Everything else, I understand your reasoning.

thank you much
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gurpreetsingh
Take the intersection of line AB and CD as point E.

EB = ED as tangents from a single point to a circle are equal

EB = EC + CD take EC = x
=> EB = x+ 24

Also AE=EC = x using same explanation above.

AB = AE+EB = x+x+24 = 2x+24 = even number- if you are short of time you can easily guess.

Now triangle ECO is similar to triangle triangle EDO' where O and O' are center of 1st and 2nd respectively.

=>\(\frac{EC}{OD} = \frac{O'D}{ED}\)

=> \(\frac{x}{5}= \frac{5}{(x+24)}\)

=> \(x^2 +24x -25 = 0\)
=> \((x +25)*(x-1)= 0\) => \(x =1\)

=> AB\(= 2x*+24 = 2+24 = 26\)


the only question I have is how did you arrive at Triangle Similarity?

we have 1 Smaller Right Triangle OCE ---- Leg = x ; Leg = 5

and 2 Larger Right Triangle EDO' ----- Leg = x + 24 ; Leg = 5

How do you find that these 2 Triangles are Similar?

Everything else, I understand your reasoning.

thank you much

I took some time to find the similarity too. Using E as the intersection point of both tangents, you will find 2 triangles EOC and PED:

1) Extend out the radius OC as a straight line onto line EP (draw it out until it touches EP, label it F)
2) See that Triangle ECF is similar to Triangle OCE; also, notice that angle CFE = angle CEO (let this be B* degrees)
3) Since all the triangles make use of a common line ED, see that angle EFC = angle EPD (corresponding angles formed from 2 parallel lines CF and DP using the line EP); this gives us the similar triangle property where angle CEO = angle EFC = angle EPD = B* degrees.
Therefore, EC/ CO = PD / ED = PD / (x+24).

Hope this helps.
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Ahhh now I see what I missed. Angle OCE and Angle PDE form 90 degree angles at the respective Points of Tangency on the SAME TANGENT LINE ED.


2 lines that form 90 degree angles with the Same Line like that will be Parallel to each other.
Therefore, Side OC is Parallel to Side DP.


Thank you. Much appreciated.

Posted from my mobile device
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I think I've figured out a Different way to Answer this question (made more intuitive sense to me at least).


Step 1: Call the Center of the Left Circle Point O and the Center of the Right Circle Point P

Connect a Straight Line from Center O to Center P.
Call the Point where this Line intersects Tangent CD - Point X


Step 2: Connect 2 Radii from Each Center to its respective Point of Tangency.

Radius drawn from Center O to A creates a 90 Degree Perpendicular Angle
Radius drawn from Center P to B creates a 90 Degree Perpendicular Angle

Since both Radii are Equal, Opposite Sides that are Parallel to each other and they create 2 Adjacent 90 degree Angles, this creates a Rectangle from Points: ABPO

If we can find the Length of Side OP, this will be Equal to the Length of AB --- which is what the Q is asking for.


Step 3:
Connect Center O to Point of Tangency C. Forms a 90 Degree Perpendicular Angle

Also Connect Center P to Point of Tangency D. Forms a 90 Degree Perpendicular Angle.

You now have 2 Triangles that are Congruent based on the A-A-A Similarity Rule and the Fact that both have an Equal Corresponding Side = Radius of 5 (OC and PD)

Triangle COX --- is Congruent to ---- Triangle DPX

Step 4:

We are told that CD = 24 meters.

Because of Point X, we have cut Line CD into Line CX and Line XD.

CX and XD are both opposite the Same Corresponding Equal Angles in the 2 Triangles. This means that the Line CD = 24 must be Bisected at Point X and Each Corresponding Leg of the 2 Congruent Right Triangles = 12

Step 5:

Lastly, Both Triangles are Congruent 5 - 12 - 13 Triangles

Hypotenuse OX = 13
Hypotenuse XP = 13

and Line OP = 26, which is Equal to and Parallel to Line AB in the Rectangle created at the outset.

Thus, Answer is:

Line AB = Line OP = 26 in

If too hard to follow and if anyone cares, I can draw a Diagram.
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