25 Points bc this depends on if I pass or fail my grade

Solve for “x”. Assume that lines which appear tangent are tangent

  48)  x = 15

  49)  x = 9

  50)  x = 17

Step-by-step explanation:

In each case, the product of lengths from the intersection point to the circle (near, far intersection points) is a constant. In the case of a tangent, the two intersection points with the circle are the same point, so the product of the lengths is the square of the length of the tangent.

48) 18^2 = 12(12 +x)

  x +12 = 18^2/12

  x = 27 -12

  x = 15

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49) 12·6 = 8·x

  x = 12·6/8

  x = 9

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50) 9·(9+21) = 10·(10 +x)

  9·30/10 = 10 +x

  x = 270/10 -10

  x = 17

- 70

this is the answer for you hope it

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy