Animated Solution for Mathematics - Conic Sections: The equation of the common tangent touching the circle $(x-3{)}^2+y^2=9$ and the parabola $y^2=4x$ above the x-axis is

We are tasked with finding a common tangent to two curves: a circle centered at $(3,0)$ with radius $3$, defined by $(x-3{)}^2+y^2=9$, and a parabola $y^2=4x$. We specifically seek the tangent line that lies in the region above the $x$-axis.

For any parabola $y^2=4ax$, the equation of a tangent with slope $m$ is given by:

By comparing our parabola $y^2=4x$ with the standard form $y^2=4ax$, we identify $4a=4$, which implies $a=1$.

Substituting $a=1$ into the slope form, we obtain the equation of our potential tangent:

This line is tangent to the parabola for any non-zero value of $m$.

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be exactly equal to the radius. Our circle $(x-3{)}^2+y^2=9$ has its center at $(3,0)$ and a radius $R=3$.

We rewrite our tangent line as $mx-y+\frac{1}{m}=0$. Using the perpendicular distance formula $d=\frac{|A{x}_0+B{y}_0+C|}{\sqrt{A^2+B^2}}$, we set the distance from $(3,0)$ to the line equal to $3$:

To solve for $m$, we square both sides of the equation to eliminate the absolute value and the radical:

Expanding the left side using the identity $(a+b{)}^2=a^2+b^2+2ab$, we get:

The $9m^2$ terms on both sides cancel out, leaving us with:

This yields two possible slopes: $m=\pm \frac{1}{\sqrt{3}}$.

The problem specifies that the tangent must be above the $x$-axis. The $y$-coordinate of the point of contact on the parabola is given by $\frac{2a}{m}=\frac{2}{m}$.

For the $y$-coordinate to be positive, $m$ must be positive. Therefore, we select $m=\frac{1}{\sqrt{3}}$.

Substituting $m=\frac{1}{\sqrt{3}}$ into the tangent equation $y=mx+\frac{1}{m}$, we get:

Multiplying the entire equation by $\sqrt{3}$ to clear the fraction, we arrive at the final result: