KVPY SX SB Exam 2012 Question Paper

Question : 4 of 120

Marks: +1, -0

Suppose two perpendicular tangents can be drawn from the origin to the circle

x^2+y^2-6x-2py+17=0

, for some real p. Then |p| is equal to

Solution:

(x − 3)^2 + (y − p)^2 = 9 -17 + p^2

Director circle is

(x-3)^2+(y-p)^2=2(p^2-8)

Passes through (0, 0)

9+p^2=2p^2-16

p^2=25

⇒p=±5⇒|p|=5

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