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!

Guest Mar 29, 2020

#1**+1 **

**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?

AnExtremelyLongName Mar 29, 2020

#2**+2 **

**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.

AnExtremelyLongName Mar 29, 2020

#3**+1 **

**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 = \(\boxed{\text{Now you can do it!}}\)

.AnExtremelyLongName Mar 29, 2020