Animated Solution for Mathematics - Circles: The line $2x-y+1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2y=4$. Then, the radius of the circle is:

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering the hidden architecture of a circle.

When you look at a problem like this, do not see just numbers and variables. See the geometry. See the elegance.

Imagine you are standing at the point of tangency, $P(2,5)$, looking out at a line that kisses the circle at that exact spot. That line is our tangent, $2x-y+1=0$. It is a boundary, a limit, a line that defines the circle's reach at that point.

The Normal

Now, here is the secret that separates the masters from the novices. Every circle has a VIP path—the normal.

The normal is the line that passes through the point of tangency and heads straight for the center. It is the radial line, and it is always, without exception, perpendicular to the tangent.

This is our golden key. If we can find the equation of this normal line, we can find the center of the circle, because the center must lie on it.

First, let us find the slope of our tangent. By rearranging $2x-y+1=0$ into the slope-intercept form, we get $y=2x+1$.

The slope, $m_{tangent}$, is clearly $2$. Since the normal is perpendicular, its slope, $m_{normal}$, must satisfy the condition $m_{normal}\cdot m_{tangent}=-1$.

Thus, $m_{normal}=-\frac{1}{2}$.

We know the normal passes through $P(2,5)$ and has a slope of $-\frac{1}{2}$. Using the point-slope form, $y-y_1=m(x-x_1)$, we write:

Let us simplify this with care. Multiplying by $2$, we get $2(y-5)=-1(x-2)$, which expands to $2y-10=-x+2$.

Rearranging, we find the equation of our normal line: $x+2y=12$. This line is the highway to the center of our circle.

Pinpointing the Center

We are told that the center of the circle also lies on the line $x-2y=4$. This is the moment of truth.

The center must satisfy both the normal line equation and this given line. We have a system of two linear equations:

1) $x+2y=12$

2) $x-2y=4$

If we add these two equations, the $2y$ and $-2y$ terms vanish in a beautiful, clean cancellation: $2x=16$, which gives us $x=8$.

Substituting $x=8$ back into $x-2y=4$, we get $8-2y=4$, leading to $2y=4$, or $y=2$. Our center, $C$, is located at $(8,2)$.

Calculating the Radius

We have arrived at the final step. The radius $r$ is simply the distance from the center $C(8,2)$ to the point of tangency $P(2,5)$.

Using the distance formula, we have:

Simplifying the surd, we get $r=3\sqrt{5}$.

Look at what we have achieved. We did not just calculate a number; we navigated the geometric properties of a circle to find its heart. Keep this clarity, keep this focus, and you will conquer any problem the JEE throws your way.