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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of normal to the circle $(x-2)^2+(y-2)^2=4$ at $(2,4)$ ?

$\begin{array}{1 1}(a)\;x=2\\(b)\;x=4\\(c)\;x=6\\(d)\;x+y=5\end{array}$

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1 Answer

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We know the equation of normal to the circle is
Hence comparing $x^2+y^2+2gx+2fy+c=0$ to
Equation (1) can be written as
Hence $g=-2,f=-2$
Now putting g & f in equ(1) we get
This is the required equation.
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v

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