$\begin{array}{1 1}(a)\;x\pm y+2=0\\(b)\;x\pm 2y+2=0\\(c)\;x\pm y+3=0\\(d)\;x\pm y+12=0\end{array}$

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For the parabola $y^2=8x$

$a=2$

Equation of tangent in slope form for parabola

$y=mx+\large\frac{2}{m}$

$y-mx-\large\frac{2}{m}$$=0$

We know if this is the common tangent that it will pass through circle hence its distance from the centre of circle will be its radius hence

$\large\frac{\bigg|0-0-\Large\frac{2}{m}\bigg|}{\sqrt{1^2+(-m)^2}}$$=5$

Squaring $\large\frac{\Large\frac{4}{m^2}}{1+m^2}$$=2$

$2=m^2(1+m^2)$

$0=m^2+m^4-2$

$(m^2+2)(m^2-1)=0$

$m^2+2\neq 0$

$m^2-1=0$

$m=\pm 1$

Hence the required tangents are $x\pm y+ 2=0$

Hence (a) is the correct answer.

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