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

Find the equation of tangent at the point $p(x_1,y_1)$ to a circle $x^2+y^2+2gx+2fy+c=0$?

$\begin{array}{1 1}(a)\;xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\\(b)\;xx_1+yy_1+g(x+x_1)-f(y+y_1)+c=0\\(c)\;xx_1+yy_1-g(x+x_1)-f(y+y_1)-c=0\\(d)\;xx_1-yy_1-g(x+x_1)-f(y+y_1)-c=0\end{array}$

1 Answer

Slope of tangent at $(x_1,y_1)$
Derivating the equation with respect to x
$2x+2yyy'+2g+2fy'+0=0$
$y'=-\large\frac{(g+x)}{f+y}$
At $(x_1,y_1)$ slope is
$y'=-\large\frac{(g+x_1)}{f+y_1}$
Equation of tangent at $(x_1,y_1)$
$(y-y_1)=-\large\frac{(g+x_1)}{(f+y_1)}$$(x-x_1)$
$(y-y_1)(f+y_1)=(g+x_1)(x_1-x)$
$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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