Suppose two perpendicular tangents can be dra | vedclass

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(C) The equation of the circle is $x^2+y^2-6x-2py+17=0$.Completing the square,we get $(x-3)^2+(y-p)^2 = 3^2+p^2-17 = p^2-8$.Since two perpendicular ta

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$5$

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(C) The equation of the circle is $x^2+y^2-6x-2py+17=0$.Completing the square,we get $(x-3)^2+(y-p)^2 = 3^2+p^2-17 = p^2-8$.Since two perpendicular tangents are drawn from the origin $(0,0)$ to the circle,the origin must lie on the director circle of the given circle.The equation of the director circle is $(x-3)^2+(y-p)^2 = 2(p^2-8)$.Since the origin $(0,0)$ lies on this circle,we substitute $x=0$ and $y=0$:$(0-3)^2+(0-p)^2 = 2(p^2-8)$$9+p^2 = 2p^2-16$$p^2 = 25$$|p| = 5$.

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

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