Q. 95.0( 2 Votes )

If the point (2, A. (- 3, - 2) (3, 4)

B. - 3, 4

C. (- ∞, - 3) (4, ∞)

D. (- ∞, - 2) (3, ∞)

Answer :

Given that point (2,k) lies outsides the circles x2 + y2 + x - 2y - 14 = 0 and x2 + y2 = 13.



We know that for a point (a, b) to lie outside the circle S, the condition to be satisfies is S11>0.


Applying S11>0 for 1st circle,


22 + k2 + 2 - 2(k) - 14>0


4 + k2 - 2k - 12>0


k2 - 2k - 8>0


k2 - 4k + 2k - 8>0


k(k - 4) + 2(k - 4)>0


(k + 2)(k - 4)>0


We know that the solution set of (x - a)(x - b)>0 for b>a is (- ∞,a)(b,∞).


The solution set for k is (- ∞, - 2)(4,∞) ..... (1)


Applying S11>0 for 2nd circle,


22 + k2 - 13>0


k2 + 4 - 13>0


k2 - 9>0


(k - 3)(k + 3)>0


The solution set for k is (- ∞, - 3)(3,∞) ...... (2)


The resultant solution set for k is the intersection of (1) and (2).


k(((- ∞, - 2)(4,∞))((- ∞, - 3)(3,∞)))


k((- ∞, - 3)(4,∞))


The correct option is (c).

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