CPhill

This is the secant-tangent theorem......it says that.....

The length of the whole secant times the length of its external segment = the square of the length of the tangent....therefore.....

12 * 5  = x^2

60 = x^2       take the square root of both sides

√60 = x

√(4*15) = x

2√15 = x

I think it's the last one, natasza....although it's a little hard  to tell. Notice that the center is (-1.5, -1)

Then, since the radius is 3, count 3 units up from the center and y =2. Count three unts down and y = -4. The last graph is the only one that has these characteristics...

Natasza...there is a theory in geometry that says when two sides of a triangle are equal, the angles opposite those sides are equal, too. Notice that in triangle AOB that OA = OB. So, that means that angle BAO = angle ABO.

Let's let the measure of each = x.  So we have,

85 + x + x =  180

85 + 2x = 180       subtract 85 from both sides

2x = 95                 divide by 2 on both sides

x = 47.5  And that's the measure of BAO. And BAO = BAC, so there's your answer.

Here's the formula, Strider

The inverse of a 3x3 matrix:

| a11 a12 a13 |-1 | a33a22-a32a23 -(a33a12-a32a13) a23a12-a22a13 | | a21 a22 a23 | = 1/DET * | -(a33a21-a31a23) a33a11-a31a13 -(a23a11-a21a13) | | a31 a32 a33 | | a32a21-a31a22 -(a32a11-a31a12) a22a11-a21a12 | with DET = a11(a33a22-a32a23)-a21(a33a12-a32a13)+a31(a23a12-a22a13)

Can you do it, now...or do you need some help??? It's really just a "plug and play" problem....but kinda' lengthy...let me know

   

x^2+y^2=5      (circle)

3x+y=7             (line)

Rearranging the second equation, we have y = 7-3x. Now, we can substitute for y in the first one....so we have

x^2 + (7 - 3x)^2 = 5       simplifying gives us

x^2 + 49 - 42x + 9x^2  = 5       subtract 5 from both sides

x^2 + 49 - 42x + 9x^2 - 5  = 0   simplify the left side, we have

10x^2 - 42x  + 44 = 0       divide by 2 on both sides

5x^2 - 21x + 22 = 0          factor

(5x -11) (x - 2)  = 0          set each factor to  0

5x - 11 = 0               x - 2 = 0

x = 11/5                   x = 2

Now we can use  y = 7-3x  to find y

So we have 

y = 7 - 3(11/5)          and    y = 7 - 3(2)

y = 7 - 33/5              and    y = 7 - 6

y = 35/5 - 33/5         and    y = 1

y = 2/5                     and y = 1

So our solutions are   (11/5 , 2/5)   and  (2 , 1)

Note that this is just the intersection of the line and the circle in two points.....

(x -10)^2  + (y+6)^2 = 36

Don't go there, zegroes......don't go there ....  !!!!

Yes....that seems to be the general consensus......

Mine, too....natasza  !!!

Happy B'Day, GL  !!!

The Beatles send a little thing along, too........