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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 - Mathematics

MCQ

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, ∞)

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Solution

(−∞, −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 \[k > 3 \text { or } k < - 3\]

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

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

Concept: Circle - Standard Equation of a Circle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Q 9 | Page 39
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