Ask Questions, Get Answers

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
At $(x_1,y_1)$ slope is
Equation of tangent at $(x_1,y_1)$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v

Related questions