Animated Solution for Mathematics - Circles: The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^2$ at the point $(2,4)$ is:

Imagine you are standing in a coordinate plane, looking at the elegant, sweeping curve of the parabola $y=x^2$. We have two points, $A(0,1)$ and $B(2,4)$, and we are tasked with finding the center of a circle that passes through both of these points while perfectly grazing the parabola at $B$.

This is not just a math problem; it is a puzzle of constraints. To solve it, we must listen to what the geometry is telling us.

First, consider the chord $AB$. Since our circle passes through both $A$ and $B$, the segment $AB$ is a chord. Geometry teaches us a powerful lesson: the center of any circle must be equidistant from the endpoints of its chords.

This means the center must lie on the perpendicular bisector of $AB$. Let us find it. The midpoint $M$ of $AB$ is:

The slope of $AB$ is $m_{AB}=\frac{4-1}{2-0}=\frac{3}{2}$. The perpendicular bisector must have a slope that is the negative reciprocal, $m_{\perp }=-\frac{2}{3}$.

Using the point-slope form, $y-\frac{5}{2}=-\frac{2}{3}(x-1)$, we simplify this to:

This is our first line of defense—the center must live somewhere on this line.

Now, let us address the tangency at $B(2,4)$. When a circle touches a curve, they share a common tangent. The radius of the circle at the point of contact is always perpendicular to this tangent. This line, perpendicular to the tangent, is the normal.

To find it, we first need the slope of the tangent to $y=x^2$ at $x=2$. The derivative is $\frac{dy}{dx}=2x$. At $x=2$, the slope $m_t=2(2)=4$.

The normal, being perpendicular, has a slope $m_n=-\frac{1}{4}$. Using the point $B(2,4)$, the equation of the normal is $y-4=-\frac{1}{4}(x-2)$, which simplifies to:

We have two lines: $4x+6y=19$ and $x+4y=18$. The center of our circle is the unique point where these two constraints meet.

To solve, we can multiply the second equation by $4$ to get $4x+16y=72$. Subtracting the first equation, $4x+6y=19$, from this gives:

Substituting this back into $x+4y=18$, we find:

The center of the circle is $C\left(-\frac{16}{5},\frac{53}{10}\right)$. We have successfully navigated the constraints, using the perpendicular bisector to handle the chord and the normal to handle the tangency. Geometry is truly a language of perfect intersections.