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

Find the equation of tangents from the point A(3,2) to the circle $x^2+y^2+4x+6y+8=0$?

$\begin{array}{1 1}(a)\;x-2y+1=0,2x-y-4=0\\(b)\;x-2y-1=0,2x+y-4=0\\(c)\;x+2y-1=0,2x-y+4=0\\(d)\;\text{None of these}\end{array}$

1 Answer

Combined equation of pair of tangents drawn from A(3,2) to the given circle $x^2+y^2+4x+6y+8=0$ can be written as
$\Rightarrow$First finding centre C(-2,-3) and radius=$\sqrt 5$
Slope of tangents from A to S=0 be m,then the equation of tangent is
The equation of tangent is $(y-2)=m(x-3)$
$mx-y+2-3m=0$-----(1)
Length of perpendicular from C(-2,-3) on (1)=$\sqrt 5$
$\large\frac{|-2m+3+2-3m|}{\sqrt{m^2+1}}$$=\sqrt 5$
$25m^2-50m+25=5m^2+5$
$20m^2-50m+20=0$
$(2m-1)(m-2)=0$
$m=\large\frac{1}{2}$ or $m=2$
Hence equation of tangents are
$x-2y+1=0,2x-y-4=0$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
 

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