Question

find the value of k for which the line y=x+2 meets the curve y^2 +(x+k)^2=2

Answer

Answer 1

Answer:

k = 0

Step-by-step explanation:

y² + (x + k)² = 2 .... (1)

y = x + 2 .... (2)

(2) ----> (1)

(x + 2)² + (x + k)² = 2

x² + 4x + 4 + x² + 2xk + k² - 2 = 0

2x² + (4 + 2k)x + (k² + 2) = 0 .... (3)

a = 2, b = 4 + 2k, c = k² + 2

D = (4 + 2k)² - 4(2)(k² + 2) = 16 + 16k + 4k² - 8k² - 16 = 16k - 8k² = 8k(2 - k)

If 8k(2 - k) ≥ 0, then (3) has roots in set of real numbers

8k(2 - k) ≥ 0 ⇒ k ≥ 0 or k ≤ 2

(k = 0) ----> (3)

x² + 2x + 1 = 0 ⇔ (x + 1)² = 0 ⇒ x = - 1

(x = - 1, k = 0) ----> (1)

y² + (- 1)² = 2 ⇒ y = ± 1

(x = - 1) ----> (2)

y = - 1 + 2 = 1

Thus, if k = 0 the line y = x + 2 and the circle y² + (x + k)² = 2 have a common point (- 1, 1)