The equation of the normal at the point (4,-1 | vedclass

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Question

Equation of circle-

$x^2+y^2-40 x+10 y=153$

$\Rightarrow(x-20)^2+(y+5)^2-25-400=153$

$\Rightarrow(x-20)^2+(y+5)^2=578$

Centre of circle $\

Vedclass · Accepted Answer

$x + 4y = 0$

Vedclass · Answer

Equation of circle-

$x^2+y^2-40 x+10 y=153$

$\Rightarrow(x-20)^2+(y+5)^2-25-400=153$

$\Rightarrow(x-20)^2+(y+5)^2=578$

Centre of circle $\equiv(20,-5)$

As we know that normal to the circle passes through its centre.

$	herefore$ Equation of normal to the circle $x ^2+ y ^2-40 x +10 y =153$ at $(4,-1)$ will be

$( y -(-1))=\frac{-5-(-1)}{20-4}( x -4)$

$\Rightarrow y +1=-\frac{1}{4}( x -4)$

$\Rightarrow 4( y +1)=4- x$

$\Rightarrow x +4 y =0$

Hence the equation of normal to the circle $x ^2+ y ^2-40 x +10 y =153$ at $(4,-1)$ is $x +4 y =0$

The equation of the normal at the point $(4,-1)$ of the circle $x^2+y^2-40 x+10 y=153$ is

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