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

Find the length of the intercept on the straight line $4x-3y-10=0$ by the circle $x^2+y^2-2x+4y-20=0$

$\begin{array}{1 1}(a)\;10&(b)\;15\\(c)\;8&(d)\;14\end{array}$

1 Answer

Equation of circle $x^2+y^2-2x+4y-20=0$
Centre and radius of the circle (1,-2) and 5
Now the distance of the point (1,-2) from the line $4x-3y-10=0$ is
$\large\frac{4(1)-(-2)\times 3-10}{5}$$=0$
Hence line passes through centre .So intercepted length =$2\times 5=10$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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