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JEE Previous Year Question

$If x^{2} + y^{2} + 6 x + 2 ky + 25 = 0 to touch Y -axis, then k =$

Topic: JEE PYQ

Answer: C. Full step-by-step solution below.

A $\pm 20$
B $- 1, - 5$
C $\pm 5$
D $4$

Correct answer: C

Solution

(c) Given circle x^{2} + y^{2} + 6 x + 2 ky + 25 = 0 touches Y -axis centre (- 3, - k) and radius = 3 units Also radius = 9 + k^{2} - 25 \Rightarrow k^{2} - 16 = 3 \Rightarrow k^{2} = 25 \Rightarrow k = \pm 5

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