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

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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.

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