# 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, ∞)

#### 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
Is there an error in this question or solution?

#### APPEARS IN

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