1. A circle centered at \(A\) with radius \(10 \) is externally tangent to a circle centered at \(B\) with radius \(7.\) A line that is externally tangent to both circles is drawn, where both circles lie on the same side of the line. This line intersects line \(AB\) at \(C\) Find the length \(BC\).
2. Lines \(PTQ\) and \(PUR\) are tangent to a circle, as shown below.
If \(\angle QTA = 41^\circ\) and \(\angle RUA = 63^\circ,\) then find \(\angle QPR,\) in degrees.
3. Points \(A\) and \(B\) lie on a circle centered at \(O\), and point \(P\) is outside the circle such that \(\overline{AP}\) and \(\overline{BP}\) are tangent to the circle. If \(\angle APO = 26^{\circ}\), then what is the measure of minor arc \(AB\), in degrees?
Thank you!
+658 Question 1:
(I drew this)
AD = 10 - 7 = 3
Through a series of proofs, we know that ADB is similar to BEC by AA.
We set up a proportion.
1. \(\frac{AD}{AB}=\frac{BE}{BC}\)
Subsitute in known values.
2. \(\frac{3}{17}=\frac{7}{BC}\)
Can you solve for BC?
+658 Question 2:
Draw:
We draw TO, OU, and OA so that they meet at the center as radii.
Plan with what you know:
Since angle QTA = 41 and angle RUA = 63, we can use that information to find angle TOU.
TO, OA, and OU are all congruent as they are radii. Because of this, isosceles triangles TOA and UOA are formed.
Solving:
PUA = 180 - RUA = 117
PTA = 180 - QTA = 139
OUA = PUA - 90 = 27
OTA = PTA - 90 = 49
UOA = 180 - 2(OUA) = 126
TOA = 180 - 2(OTA) = 82
TOU = 360 - (UOA + TOA) = 152
QPR = 360 - 90 - 90 - TOU = \(\boxed{\text{Now you do the rest!}}\)
Shouldn't be that hard, considering that you read my solution.
+658 Question 3:
(I drew this)
Consider:
Angle APO = 26 and the right angles formed by tangent and radii lines
By the Two-Tangent theorem, BP = PA
By a serious of proofs, we know that triangles BPO and APO are congruent by SSS.
Solve:
Because of CPCTC, we know that BPO is congruent to APO and is therefore also 26 degrees.
We solve for BOP and AOP
BOP = 180 - 26 - 90 = 64
AOP = 180 - 26 - 90 = 64
BOA = 64 + 64
Minor arc AB = \(\boxed{\text{Now you can do it!}}\)
