Question bank/JEE Main/Mathematics

JEE MainMathematicsEasyCircles

If two perpendicular tangents can be drawn from the origin to
the circle
x2−6x+y2−2py+17=0x^{2} - 6x + y^{2} - 2py + 17 = 0then the value of p2p^{2}
is;fn ewy fcUnq ls o`r
x2−6x+y2−2py+17=0 x^{2} - 6x + y^{2} - 2py + 17 = 0\ ij nks

Show answer and solution
(x−3)2+(y−p)2=p2−8(x - 3)^{2} + (y - p)^{2} = p^{2} - 8
(0, 0) must lie on the director circle:

(x−3)2+(y−p)2=2(p2−8)(x - 3)^{2} + (y - p)^{2} = 2\left( p^{2} - 8 \right)
(0, 0) fu;Xed o`r:

(x−3)2+(y−p)2=2(p2−8)(x - 3)^{2} + (y - p)^{2} = 2\left( p^{2} - 8 \right)ij fLFkr gksxkA
p2=25p^{2} = 25

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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