Question

If the point (2, k) lies outside the circles x² + y² + x − 2y − 14 = 0 and x² + y² = 13 then k lies in the interval

Options

• (−3, −2) ∪ (3, 4)

• −3, 4

• (−∞, −3) ∪ (4, ∞)

• (−∞, −2) ∪ (3, ∞)

Answer 1

(−∞, −3) ∪ (4, ∞)

The given equations of the circles are x² + y² + x − 2y − 14 = 0 and x² + y² = 13.
Since (2, k) lies outside the given circles, we have: \[4 + k^2 + 2 - 2k - 14 > 0\] and \[4 + k^2 > 13\]

\[\Rightarrow k^2 - 2k - 8 > 0\] and  \[k^2 > 9\]

\[\Rightarrow \left( k - 4 \right)\left( k + 2 \right) > 0\] and  \[k^2 > 9\]

\[\Rightarrow k > 4 \text { or } k < - 2\]  and \[k > 3 \text { or } k < - 3\]

\[\Rightarrow k > 4 \text { and } k < - 3\]

\[\Rightarrow k \in \left( - \infty , - 3 \right) \cup \left( 4, \infty \right)\]