For which value(s) of the constant k is the circle x² + (y − k)² = 16 tangent to the line y = 3?

-1, 7

Step-by-step explanation:

Equation of the circle:

x² + (y − k)² = 16

When the circle intersects y = 3:

x² + (3 − k)² = 16

x² + 9 − 6k + k² = 16

x² = 7 + 6k − k²

x = ±√(7 + 6k − k²)

For the circle to be tangent to the line, it can only intersect at one point.  If x has only one value, then:

√(7 + 6k − k²) = -√(7 + 6k − k²)

2√(7 + 6k − k²) = 0

7 + 6k − k² = 0

k² − 6k − 7 = 0

(k − 7) (k + 1) = 0

k = -1, 7

The two values of k are -1 and 7.

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.

answer

step-by-step explanation: