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10-1.Circle and System of Circles
hard
Suppose two perpendicular tangents can be drawn from the origin to the circle $x^2+y^2-6 x-2 p y+17=0$, for some real $p$. Then, $|p|$ is equal to
A
$0$
B
$3$
C
$5$
D
$17$
(KVPY-2012)
Solution
(c)
We have,
Equation of circle
$x^2+y^2-6 x-2 p y+17 =0$
$\Rightarrow \quad(x-3)^2+(y-p)^2 =p^2-8$
Given two perpendicular tangents drawn from origin to the circle.
$\therefore$ Locus is director circle of the given circle.
$\therefore$ Equation of director circle is
$(x-3)^2+(y-p)^2=2\left(p^2-8\right)$
This passes through $(0,0)$.
$\therefore 9+p^2=2 p^2-16$
$\Rightarrow p^2=25 \Rightarrow p=\pm 5$
$\therefore |p|=|\pm 5|=5$
Standard 11
Mathematics
