The equation of the common tangent to the cir | vedclass

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(D) The given parabola is $y^{2}=8x$. Comparing with $y^{2}=4ax$,we get $4a=8$,so $a=2$.Any tangent to the parabola is given by $y=mx+\frac{a}{m}$,whi

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$-2$

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(D) The given parabola is $y^{2}=8x$. Comparing with $y^{2}=4ax$,we get $4a=8$,so $a=2$.Any tangent to the parabola is given by $y=mx+\frac{a}{m}$,which can be written as $mx-y+\frac{2}{m}=0$.For this line to be a tangent to the circle $x^{2}+y^{2}=2$ (with center $(0,0)$ and radius $r=\sqrt{2}$),the perpendicular distance from the center to the line must equal the radius.Therefore,$\frac{|\frac{2}{m}|}{\sqrt{m^{2}+1}}=\sqrt{2}$.Squaring both sides,$\frac{4}{m^{2}}=2(m^{2}+1)$ $\Rightarrow 2=m^{2}(m^{2}+1)$ $\Rightarrow m^{4}+m^{2}-2=0$.Let $t=m^{2}$,then $t^{2}+t-2=0 \Rightarrow (t+2)(t-1)=0$. Since $m^{2} \geq 0$,we have $m^{2}=1$,so $m=\pm 1$.If $m=1$,the tangent is $x-y+2=0 \Rightarrow y=x+2$.If $m=-1$,the tangent is $-x-y-2=0 \Rightarrow x+y=-2$.Comparing $x+y=-2$ with the given form $x+y=k$,we get $k=-2$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

The equation of the common tangent to the circle $x^{2}+y^{2}=2$ and the parabola $y^{2}=8x$ is $x+y=k$. Then the value of $k$ is

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