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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equation of normal for the circle, $x^2+y^2+2gx+2fy+c=0$ at $(x_1,y_1)$

$\begin{array}{1 1}(a)\;\large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)}\\(b)\;\large\frac{(y-y_1)}{(f-y_1)}=\frac{(x+x_1)}{(g-x_1)}\\(c)\;\large\frac{(2y-y_1)}{(f+y_1)}=\frac{(2x-x_1)}{(g+x_1)}\\(d)\;\text{None of these}\end{array}$

1 Answer

We know slope of tangent =$-\large\frac{(g+x_1)}{(f+y_1)}$
Hence slope of normal =$-\large\frac{f+y_1}{g+x_1}$
Hence equation of normal
Hence (a) is the correct answer.
Given equation of the circle is
Differentiating both the sides with respect to $x$ we get
Slope of tangent at any point $(x_1,y_1)$ to any curve is
$\large\frac{dy}{dx}$ at $(x_1,y_1)$
$i.e.,$ Slope of tangent is - $\large\frac{x_1+g}{y_1+f}$
Normal is $\perp$ to tangent
$\therefore$ Slope of normal is $-\large\frac{1}{dy/dx}$
$\therefore$ Equation of normal at $(x_1,y_1)$ is
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
edited Mar 19, 2014 by rvidyagovindarajan_1

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