Two circles are externally tangent at T. The line AB is a common external tangent to the two circles, and P is the foot of the altitude from T to line AB. Find the length AB.
We can set variables to complete this problem quite efiiciently.
First, let's set the center of the large circle as M and the center of the smaller circle as N.
Now, let's extend AB and and the line connecting the centers until they meet each other at a pont, O.
Let the distance of the left edge of the smaller circle and O be x.
Now, let's note that triangle BMO and ANO are similar. This is quite important.
From this, we can write
BM/MO=AN/NO4/(4+2+x)=1/(1+x)4/(6+x)=1/(1+x)4(1+x)=1(6+x)4+4x=6+x3x=2x=2/3
Now that we have the value for x, we can easily solve for AB. We get that
OA=√NO2−NA2=√(1+2/3)2−12=√25/9−1=√16/9=4/3OB=√MO2−MB2=√(6+2/3)2−42=√400/9−16=√256/9=16/3AB=OB=OA=16/3−4/3=12/3=4
So our answer is 4
Feel free to let me know if I messed up!
Thanks! :)
We can set variables to complete this problem quite efiiciently.
First, let's set the center of the large circle as M and the center of the smaller circle as N.
Now, let's extend AB and and the line connecting the centers until they meet each other at a pont, O.
Let the distance of the left edge of the smaller circle and O be x.
Now, let's note that triangle BMO and ANO are similar. This is quite important.
From this, we can write
BM/MO=AN/NO4/(4+2+x)=1/(1+x)4/(6+x)=1/(1+x)4(1+x)=1(6+x)4+4x=6+x3x=2x=2/3
Now that we have the value for x, we can easily solve for AB. We get that
OA=√NO2−NA2=√(1+2/3)2−12=√25/9−1=√16/9=4/3OB=√MO2−MB2=√(6+2/3)2−42=√400/9−16=√256/9=16/3AB=OB=OA=16/3−4/3=12/3=4
So our answer is 4
Feel free to let me know if I messed up!
Thanks! :)