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

If the tangent at the point P on the circle $x^2+y^2+6x=2$ meets a straight line $5x-2y+6=0$ at a point Q on the y-axis , then the length of $PQ$ is

$\begin{array}{1 1}(A)\;4\\(B)\;2 \sqrt 5\\(C)\;5 \\(D)\;3 \sqrt 5 \end{array}$

1 Answer

Line $5x-2y+6=0$ is intersected by tangent P to circle $x^2+y^2+6x+6y-2=0$ on y-axis at $Q(0,3)$
In other words tangent passes through $(0,3)$
$PQ$= length of tangent to circle from $(0,3)$
$\qquad= \sqrt {0+9+0+18-2}$
$\qquad= \sqrt {25}$
$\qquad= 5$
Hence C is the correct answer.
answered Apr 8, 2014 by meena.p
 

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