CPhill

Thx, jugoslav....!!!!

We have the points  (4,-7)  and (0, -6)  on the  graph

Slope    of the line through  these ponts   =     [ -6 - - 7 ]  /  [ 0 -4 ] =    -1/4    =  m

We have the form

y = mx  +  b

Where  m  = the slope   b =  the y intercept  =  -6

So we have that

y  = (-1/4)x  - 6

If the line of symmetry  is  x =2.....then this is the x coordinate of the  vertex

We have the form

y =  a(x - h)^2  + k         where  h  is the  x oordinate of the vertex ....we need to find a  and k thusly

1  = a  (1 - 2)^2  +  k   →  1  = a  +  k

-1  = a  (5 - 2)^2  + k  → -1  = 9a + k

Subtract  the second equation from the first

2  = -8a

a = 2/-8  = -1/4

And   1 =  -1/4  +  k

k  = 1 + 1/4  = 5/4

So  we  have that

y  =(-1/4) ( x - 2)^2 +  5/4

y = (-1/4)  ( x^2  -4x  + 4) + 5/4

y = (-1/4)x^2  + x  -1  +  5/4

y  = (-1/4)x^2  + x  + 1/4

a = -1/4    b  =  1     c   =   1/4

And their sum is     1

arcsin  (-0.31)  ≈  -18°  = x

990/360   =  2.75  =  2 + 3/4   rotations

We  will have  two angles in each rotation  where  sin  x  = -0.31

360 - 18  = 342°

360 (3/4)  = 270°

So....in the last rotation, we only have  one  angle value  where sin x = -0.31

So....   2(2)  + 1  =   5  values of  x

Area of  right triangle   =(1/2)  (product of legs)  = (1/2) (3 * 4)  =  (1/2)(12)  =6

Area of  triangle   AGC  = (1/2) AC * GC  = (1/2) (5) (1/2)  = 5/4

Area of  triangle  BGC  =  (1/2) BC * GP  = (1/2) ( 4) (1)  = 2

Area of triangle  AGB =  6 - 5/4  - 2  =  4  - 5/4   =   11/4

Area of triangle AGB  = (1/2) AB * RG 

11/4 =  (1/2) 3 * RG

11/4 = 3/2 * RG

RG  = (11/4) (2/3)  =  22/12  =  11/6   =  distance  from G to AB

f(0)  = a/2

To find the inverse :

y =  a  / (x + 2)         get x by itself

y ( x + 2)  = a

xy  + 2y  =  a

xy  = a - 2y  

x =  (a -2y)  / y   swap x and  y

y = (a - 2x ) / x    =  the inverse to f(x)

f^-1 (3a)  =   (a -2(3a) ) / (3a)  =  -5a/ 3a   =  -5/3

Therefore

f(0)  = f^-1 (3a)    gives us

a/2  =  -5/3

a = 2 (-5/3)  =  -10/3

By  the intersecting  chord theorem

PQ * CP  = PB * AP

PB  = r + 25

AP  = r -25

So

8 * CP = (r + 25) (r -25)

CP  =  (r^2 -625)  / 8      →  CP^2  =  (r^2 -625)^2  / 64

Therefore

CP^2 - PO^2  =  CO^2  = r^2 

(r^2 - 625)^2 /64  -  25^2  =  r^2

r^4 - 1250r^2  + 390625  -  40000  =  64r^2

r^4  - 1314r^2  + 350625  =  0

Solving  this  produces  that  the possible positive values for  r   ≈  19.296   or  r ≈ 30.686

r  must  be > 25   ....so....r ≈  30.686

We can use  the tangent-secant theorem to find   CB

We  have  that

AB^2  = CB  *( CB + DC)

15^2  =  CB ( CB  + 10)

225 = CB^2  + 10CB

CB^2  + 10CB  -225  =  0

Solving this quadratc (taking the positive result) gives us  that CB  =   5sqrt (10)  - 5

So  the  radius =  (1/2)DC  + CB   =   (1/2) 10  +  5sqrt (10)  - 5  =

5 + 5sqrt (10)   - 5   = 

5sqrt (10)

We can use the  secant-secant  product theorem  to find PC

AC (AC + PC)  = AB (AB + SB)

8 ( 8 + PC)  = 5 (5 + 25)

64 + 8PC  =  5 (30)

8PC =  150 - 64

8PC = 86

PC  =  86/8  =  43/4  =10.75

So   AP  = 8 + 10.75  = 18.75

So  SP^2  =   [ AS^2 - AP^2 ] =   [ 30^2  - 18.75^2] 

Connect   SC  and we  have right triangle  SCP....and since APS is a right angle....then SC will be the diameter of the  circle

So  SC   = sqrt  [ PC^2  + SP^2 ]  =  sqrt  [ 10.75^2  + 30^2  - 18.75^2  ]   = sqrt (664)

So....the radius =  sqrt (664) / 2

And the area of the  circle is     (sqrt (664) / 2 )^2 * pi  =   (664/4)  pi = 166 pi  units^2

Since the circle is tangent at (0,5)   the  center  is  (a,5)

The  distance  from  this point to the gfiven points is equal.....so we can solve the simplified distance^2  formula for  a

a^2   =   (12 -a)^2  + ( 5)^2    simplify

a^2 = a^2 - 24a  + 144 + 25

24a = 169

a = 169/24    =  the  radius

See  the graph here :  https://www.desmos.com/calculator/tm67fzywd1