Animated Solution for Mathematics - Conic Sections: Comprehension Passage

First, let's look at our circle: $x^2+y^2-8x=0$. By completing the square, we rewrite it as $(x-4{)}^2+y^2=16$.

We identify a circle centered at $(4,0)$ with a radius of $4$. It is a symmetrical shape sitting on the $x$-axis.

Now, consider the hyperbola: This is the standard form of a hyperbola opening along the $x$-axis. Our goal is to find a line that acts as a common tangent to both curves.

To find a common tangent, we assume the equation of a line with slope $m$ is $y=mx+c$. For this line to be tangent to the hyperbola, it must satisfy the condition $c^2=a^2{m}^2-b^2$.

Given $a^2=9$ and $b^2=4$, the condition becomes $c^2=9m^2-4$. Thus, $c=\pm \sqrt{9m^2-4}$.

For this line to also touch the circle, the perpendicular distance from the center $(4,0)$ to the line $mx-y+c=0$ must equal the radius $4$. Using the distance formula, we set:

Squaring both sides and substituting $c^2=9m^2-4$, we solve for $m$. After algebraic manipulation, we arrive at $495m^4+104m^2-400=0$. Solving for $m^2$, we find $m^2=\frac{4}{5}$, leading to the tangent equation: $2x-\sqrt{5}y+4=0$.

Next, we find the intersection points $A$ and $B$ of the circle and the hyperbola. We substitute $y^2=8x-x^2$ from the circle into the hyperbola equation:

Multiplying by $36$ to clear the fractions, we obtain $4x^2-9(8x-x^2)=36$, which simplifies to $13x^2-72x-36=0$. Factoring this quadratic yields $(x-6)(13x+6)=0$.

We discard the negative root as the intersection occurs on the right branch. Thus, $x=6$. Substituting this into the circle equation gives $y^2=12$, so $y=\pm 2\sqrt{3}$. The intersection points are $A(6,2\sqrt{3})$ and $B(6,-2\sqrt{3})$.

We now determine the equation of the circle with $AB$ as its diameter. Using the diameter form $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$, we substitute our points:

Expanding this expression, we get $(x-6{)}^2+(y^2-12)=0$. Simplifying further, we arrive at the final equation:

This result represents the circle perfectly defined by the intersection of the original two curves. Through this process, we have successfully navigated the geometry of these conics.