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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
$\large\frac{y-y_1}{f+y_1}=\frac{x-x_1}{(g+x_1)}$-------(1)
Hence comparing $x^2+y^2+2gx+2fy+c=0$ to
$(x-2)^2+(y-2)^2=4$------(2)
Equation (1) can be written as
$x^2+y^2-4x-4y+4=0$
Hence $g=-2,f=-2$
Now putting g & f in equ(1) we get
$\large\frac{y-4}{-2+4}=\frac{x-2}{(-2+2)}$
$\large\frac{y-4}{2}=\frac{x-2}{(0)}$
$x=2$
This is the required equation.
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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