Animated Solution for Mathematics - Circles: A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point :

Imagine you are standing on a vast, flat coordinate plane. You are tasked with drawing a circle that kisses the x-axis at exactly one point: $(3,0)$.

Because the circle touches the x-axis at $(3,0)$, the tangent at this point is horizontal. A fundamental rule of circles tells us that the radius drawn to the point of tangency is always perpendicular to the tangent.

Since our tangent is the x-axis, our radius must be a vertical line. This forces the center of our circle to lie on the vertical line $x=3$. Let us denote the center as $C(3,k)$. The radius $r$ is the distance from this center to the x-axis, which is simply the vertical distance from the x-axis to the center, giving us $r=|k|$.

Now, the problem introduces a second constraint: the circle makes an intercept of length $8$ on the y-axis. Visualize the y-axis as a vertical wall. The circle cuts through this wall, creating a chord of length $8$.

Here, we invoke a beautiful circle theorem: a perpendicular drawn from the center of a circle to a chord bisects that chord. If we drop a perpendicular from our center $C(3,k)$ to the y-axis, the length of this perpendicular is the horizontal distance from the center to the y-axis, which is exactly $3$ units.

This perpendicular bisects our $8$-unit chord into two equal segments of $4$ units each.

We are now standing on the precipice of the solution. If we connect the center $C(3,k)$ to one of the endpoints of the chord on the y-axis, we form a right-angled triangle.

The base of this triangle is the perpendicular distance to the y-axis, which is $3$. The height is half the chord length, which is $4$. The hypotenuse is the radius $r$.

By the Pythagorean theorem, we have:

Calculating this, we find $r^2=9+16=25$, which means $r=5$. Since $r=|k|$, we have $|k|=5$, leading to $k=5$ or $k=-5$. Given the context of the options, we select $k=5$, placing our center at $C(3,5)$.

With the center $(3,5)$ and radius $r=5$ in hand, we can construct the equation of our circle using the standard form:

Substituting our values, we get:

Now, we test the options. For the point $(3,10)$, we substitute $x=3$ and $y=10$ into our equation:

The equation holds true! The point $(3,10)$ lies perfectly on the circle. We have navigated the constraints, applied the theorems, and arrived at the truth. The final result is the circle defined by $(x-3{)}^2+(y-5{)}^2=25$.