If the common tangent to the parabolas y^{2}= | vedclass

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(C) The equation of a tangent to the parabola $y^{2}=4x$ with slope $m$ is $y=mx+\frac{1}{m}$.The equation of a tangent to the parabola $x^{2}=4y$ wit

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$1/\sqrt{2}$

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(C) The equation of a tangent to the parabola $y^{2}=4x$ with slope $m$ is $y=mx+\frac{1}{m}$.The equation of a tangent to the parabola $x^{2}=4y$ with slope $m$ is $y=mx-m^{2}$.For the common tangent,we equate the intercepts: $\frac{1}{m}=-m^{2}$,which implies $m^{3}=-1$,so $m=-1$.Substituting $m=-1$ into the tangent equation,we get $y=-x-1$,or $x+y+1=0$.This line touches the circle $x^{2}+y^{2}=c^{2}$. The perpendicular distance from the center $(0,0)$ to the line $x+y+1=0$ must equal the radius $c$.Using the formula $d=\frac{|ax_{0}+by_{0}+k|}{\sqrt{a^{2}+b^{2}}}$,we have $c=\frac{|0+0+1|}{\sqrt{1^{2}+1^{2}}}=\frac{1}{\sqrt{2}}$.

If the common tangent to the parabolas $y^{2}=4x$ and $x^{2}=4y$ also touches the circle $x^{2}+y^{2}=c^{2}$,then $c$ is equal to

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