PLS HELP, DUE TMROW

1)

Circle O is a unit circle. Segment AS has length 12/5 and is tangent to circle O at A. If P is the intersection of OS with circle O, find length PS.

AS  = 12/5      AO  = 1

So   ....  

SO  =  sqrt [ (12/5)^2  + 1 ]  =   sqrt [  144 + 25] / 5  =  13/5

So...PS   =   SO  - PO  =   13/5  -  1  =     8/5

2)

Angle A : Angle P : Angle ASP are in ratio 1 : 2 : 2. Find the degree measure of angle BSA.

Angle A  =  36°......Angle P, ASP  =  72°

Angle  ASP  =  (1/2)minor arc AS  =  angle BSA  =  72°

3)

If angle B = 39 degrees and arc PS = 116 degrees, find the degree measure of arc AS.

Angle B =  (1/2) ( arc AS - arc AP)

38  =  (1/2) (arc AS - arc AP)

76  =  arc AS - arc AP     (1)

And

arc AS + arc AP  + arc PS  = 360

arc AS + arc AP   +  116  =  360

arc AS  + arc AP  =  244   (2)

Add  (1)  and (2)

arc AS - arc AP   =  76

arc AS +  arc AP  = 244

2 arc AS  =  320      divide by 2

arc AS  =  160°

4)

Points A and B are on a circle centered at O, and point P is outside the circle such that PA and PB are tangent to the circle. If angle OPA = 32 degrees, then what is the measure of minor arc AB, in degrees?

Draw radii OA, OB...so  OAPB forms a quadrilaterlal....the sum of its interior angles = 360°

Angles OAP, OBP  =  90°  and OPA  =  32°, then angle BPA  =  64°

So...angle OAB  =  360 - 2(90) - 64  =  116°

And  OAB  is a central angle  intercepting minor arc AB, so its measure is also 116°

5)

Circle O and circle P, with radii 3 and 5, respectively, are both tangent to line L at H. Enter all possible lengths of OP separated by commas.

{Need a pic, here  }

6)

Given regular pentagon ABCDE, a circle can be drawn that is tangent to DC at D and to AB at A. What is the number of degrees in minor arc AD?

Call the center of the circle O, connect  OA  and OD

And OABCD  forms another pentagon whose interior angles sum to 540°

Angles  ODC and angle OAB   =  90°

Angles DCB and CBA  =  108°

So  angle DOA  =  540 - 2(90) - 2(108)  = 144°  ...this is a central angle in the circle intercepting minor arc  AD  ....so it also measures 144°

8)

In triangle ABC, AB = 6, BC = 8  and CA = 10. A circle centered at B is tangent to AC. What is the radius of this circle?

Here's one way to do this: 

Let  B = (0,0), A  = (0,6)  and C  = (8,0)

The side  AC  has a slope  =  -6/ 8  =  -3/4

And the equation of this line  is

y = (-3/4)x + 6

We need to  find a line perpendicular to this passing through (0,0)

The slope of this line is  4/3     and its equation is:

y  = (4/3)x

To find out where these two lines intersect we have

(-3/4)x + 6  =  (4/3)x

6  =  (4/3  + 3/4) x

6 = ( 25/12) x

(12 * 6) / 25  = x

72/25  = x

And   y  = (4/3) (72/25)   =  96/25

So  the radius of a circle centered at B and tangent to AC  is

sqrt  [ (72/25)^2  + (96/25^2 ]  = sqrt  [ 72^2  + 96^2 ] /25  =

sqrt [ 14400] / 25 =

120/25  =

24/5  

11)

Two circles, centered at A and B are externally tangent to each other, and tangent to a line L. A third circle, centered at C is externally tangent to the first two circles, and the line L. If the radii of circle A and circle B are 9 and 16, respectively, then what is the radius of circle C?

Picture: https://latex.artofproblemsolving.com/4/d/a/4da24a4edb9693476d2766d290f52dd917498a6f.png

\(\text{Pythagorean theorem:} \)

\(\begin{array}{|lrcll|} \hline (1) & (r_a+r_b)^2 &=& (r_b-r_a)^2 + s_1^2 \\ & r_a^2+2r_ar_b+r_b^2 &=& r_b^2-2r_br_a+r_a^2+s_1^2 \\ & 4r_ar_b & = & s_1^2 \\ & \mathbf{s_1} &\mathbf{=}& \mathbf{2\sqrt{r_ar_b}} \\\\ (2) & (r_b+r_c)^2 &=& (r_b-r_c)^2 + s_3^2 \\ & r_b^2+2r_br_c+r_c^2 &=& r_b^2-2r_br_c+r_c^2 + s_3^2 \\ & 4r_br_c &= & s_3^2 \\ & \mathbf{s_3} &\mathbf{=} & \mathbf{2\sqrt{r_br_c}} \\\\ (3) & (r_a+r_c)^2 &=& (r_a-r_c)^2 + s_2^2 \\ & r_a^2+2r_ar_c+r_c^2 &=& r_a^2-2r_ar_c+r_c^2 +s_2^2 \\ & 4r_ar_c & = & s_2^2 \\ & \mathbf{s_2} &\mathbf{=} & \mathbf{2\sqrt{r_ar_c}} \\\\ & \mathbf{s_1} &\mathbf{=}& \mathbf{s_2 + s_3} \\ & 2\sqrt{r_ar_b} &=& 2\sqrt{r_ar_c} + 2\sqrt{r_br_c} \quad & | \quad : 2 \\ & \sqrt{r_ar_b} &=& \sqrt{r_ar_c} + \sqrt{r_br_c} \\ & \sqrt{r_ar_b} &=& \sqrt{r_c}(\sqrt{r_a} + \sqrt{r_b}) \\ & \sqrt{r_c} &=& \dfrac{\sqrt{r_ar_b}} {\sqrt{r_a} + \sqrt{r_b}} \quad & | \quad \Rightarrow \frac{1}{\sqrt{r_c}} = \frac{1}{\sqrt{r_a}} + \frac{1}{\sqrt{r_b}} \\ & r_c &=& \dfrac{ r_ar_b } {(\sqrt{r_a} + \sqrt{r_b})^2} \quad & | \quad r_a = 9 \quad r_b=16 \\ & r_c &=& \dfrac{ 9\cdot 16 } {(\sqrt{9} + \sqrt{16})^2} \\ & r_c &=& \dfrac{ 144 } {(3+4)^2} \\ & r_c &=& \dfrac{ 144 } {49} \\ \hline \end{array}\)

The radius of circle C is \(\mathbf{\tfrac{144}{49}} \)

9)

Each side of quadrilateral ABCD is tangent to a circle. If AB = 84, BC = 71 and CD = 75, find DA.

Call  the two tangents that meet at A, x

Call the two tangents that meet at B, y

Call the two tangents that meet at C, z

Call the two tangents that meet at D, w

And we have the following system

x + y   =  84     ⇒    y   =  84  - x

y + z  =   71  ⇒   z = 71 - y ⇒    z  =  71  - (84- x)  ⇒  x - 13

z + w  = 75  ⇒   w  = 75 - z  ⇒  75 - (x - 13)  =  88 - x

w + x  =    (88 - x) + x   =  88   =   DA

10)

Triangle ABC, inscribed in a circle, has AB = 15 and BC = 25. A tangent to the circle is drawn at B, and a line through A parallel to this tangent intersects BC at D. Find DC. 

Picture: https://latex.artofproblemsolving.com/e/f/6/ef688cf1577b8cd6d2c7da6f4b56dab5dee8e935.png

Let \(DC=x\)

Let \(BD = 25-x\)

Let O the centre of the circle.

Let r the radius of the circle.

\(\begin{array}{|rcll|} \hline \dfrac {\sin(\alpha)}{25-x} &=& \dfrac {\sin(\beta)}{15} \quad & | \quad \cos(90^{\circ}-\alpha) = \dfrac{15}{2r}\quad \cos(90^{\circ}-\beta) = \dfrac{25}{2r} \\ &&\quad & | \quad \dfrac{\sin(\alpha)}{15} = \dfrac{\sin(\beta)}{25} \\ &&\quad & | \quad \dfrac{\sin(\alpha)}{\sin(\beta)} = \dfrac{15}{25} \\ &&\quad & | \quad \dfrac{\sin(\alpha)}{\sin(\beta)} = \dfrac{3}{5} \\ \dfrac{\sin(\alpha)}{\sin(\beta)} &=& \dfrac{25-x} {15} \quad & = \quad\dfrac{3}{5} \\\\ \dfrac{25-x} {15} & = & \quad\dfrac{3}{5} \\\\ 5(25-x) & = & 45 \\ 25-x & = & 9 \\ x & = & 25-9 \\ \mathbf{ x } & \mathbf{=} & \mathbf{16} \\ \hline \end{array}\)

DC is 16