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

Find the equation of chord of contact of tangent drawn from a point $(x_1,y_1)$ to the circle $x^2+y^2=a^2$?

$\begin{array}{1 1}(a)\;xx_1+yy_1=a^2\\(b)\;xx_1-yy_1=a^2\\(c)\;xx_1+yy_1=a\\(d)\;xx_1^2+yy_1^2=a^2\end{array}$

1 Answer

Let $T(x',y')$ and $T'(x'',y'')$ be the points of contacts of tangents drawn from $P(x_1,y_1)$ to $x^2+y^2=a^2$
Then equation of PT and PT' are $xx'+yy'=a^2$
$xx''+yy''=a^2$
Since both tangents pass through $(x_1,y_1)$ then
$x_1x'+y_1y'=a^2$
$x_1x''+y_1y''=a^2$
Point T and T' lie on
$xx_1+yy_1=a^2$
$\therefore$ Equation of chord of contact TT' is $xx_1+yy_1=a^2$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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