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(D) The equation of a tangent to the parabola $y^2=x$ with slope $m$ is given by $y = mx + \frac{1}{4m}$.For this to be a tangent to the circle $x^2+y

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$x-2y+1=0$

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(D) The equation of a tangent to the parabola $y^2=x$ with slope $m$ is given by $y = mx + \frac{1}{4m}$.For this to be a tangent to the circle $x^2+y^2-6y+4=0$,the perpendicular distance from the center $(0, 3)$ to the line $mx - y + \frac{1}{4m} = 0$ must equal the radius of the circle.The circle equation can be written as $x^2 + (y-3)^2 = 5$,so the radius is $\sqrt{5}$.Thus,$\frac{|m(0) - 3 + \frac{1}{4m}|}{\sqrt{m^2 + 1}} = \sqrt{5}$.Squaring both sides: $\frac{(-3 + \frac{1}{4m})^2}{m^2 + 1} = 5$.$9 - \frac{3}{2m} + \frac{1}{16m^2} = 5m^2 + 5$.Multiplying by $16m^2$: $144m^2 - 24m + 1 = 80m^4 + 80m^2$.$80m^4 - 64m^2 + 24m - 1 = 0$.Testing $m = 1/2$: $80(1/16) - 64(1/4) + 24(1/2) - 1 = 5 - 16 + 12 - 1 = 0$.So,$m = 1/2$ is a solution.The equation of the tangent is $y = \frac{1}{2}x + \frac{1}{4(1/2)} = \frac{1}{2}x + \frac{1}{2}$.Multiplying by $2$: $2y = x + 1$,which simplifies to $x - 2y + 1 = 0$.

The equation of one of the common tangents of the circle $x^2+y^2-6y+4=0$ and the parabola $y^2=x$ is

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