Let C₁ be the circle of radius 1 with center at the origin. Let C₂ be the circle of radius r with center at the point A = (4, 1), where 1 < r < 3. Two distinct common tangents PQ and ST of C₁ and C₂ are drawn. The tangent PQ touches C₁ at P and C₂ at Q. The tangent ST touches C₁ at S and C₂ at T. Midpoints of the line segments PQ and ST are joined to form a line which meets the x-axis at a point B. If AB = $\sqrt{5}$, then the value of r² is:

The correct answer is 2.

Let us analyze the problem step-by-step.

Step 1: Understand the given equations of the circles
Circle C₁ has its center at the origin (0, 0) and radius R₁ = 1.
The equation of C₁ is:

$x^2+y^2=1$

Or in expanded form:

$S_1\equiv x^2+y^2-1=0$

Circle C₂ has its center at A = (4, 1) and radius R₂ = r.
The equation of C₂ is:

${(x-4)}^2+{(y-1)}^2=r^2$

Expanding this equation:

$x^2-8x+16+y^2-2y+1=r^2$

$S_2\equiv x^2+y^2-8x-2y+17-r^2=0$

Step 2: Understand the locus of midpoints of common tangents
Let M be the midpoint of the line segment PQ (a common tangent to C₁ and C₂).
The length of the tangent from M to C₁ is equal to the length of the tangent from M to C₂ because M is the midpoint of PQ (so MP = MQ).
By definition, the locus of points from which the lengths of tangents to two circles are equal is the radical axis of the two circles.
Since M is the midpoint of PQ and similarly the midpoint of ST lies on the same locus, the line joining these midpoints is the radical axis of circle C₁ and circle C₂.

Step 3: Find the equation of the radical axis
The equation of the radical axis of two circles S₁ = 0 and S₂ = 0 is given by:

$S_1-S_2=0$

Subtracting the equation of S₂ from S₁:

$(x^2+y^2-1)-(x^2+y^2-8x-2y+17-r^2)=0$

$8x+2y-18+r^2=0$

Step 4: Find the coordinates of point B
Point B is the intersection of the radical axis with the x-axis.
Setting y = 0 in the equation of the radical axis:

$8x+2\left(0\right)-18+r^2=0$

$8x=18-r^2$

$x=\frac{18-r^2}{8}$

Thus, the coordinates of point B are:

$B=(\frac{18-r^2}{8},0)$

Step 5: Calculate distance AB and solve for r²
We are given that A = (4, 1) and AB = $\sqrt{5}$.
Using the distance formula:

$A{B}^2={(4-x_B)}^2+{(1-0)}^2=5$

${(4-x_B)}^2+1=5$

${(4-x_B)}^2=4$

Taking the square root on both sides:

$4-x_B=2$ or $4-x_B=-2$

This gives:

$x_B=2$ or $x_B=6$

Now substitute $x_B=\frac{18-r^2}{8}$:

Case 1: If x_B = 2

$\frac{18-r^2}{8}=2$

$18-r^2=16$

$r^2=2$

$r=\sqrt{2}$

Case 2: If x_B = 6

$\frac{18-r^2}{8}=6$

$18-r^2=48$

$r^2=-30$ (which is not possible for real radius r)

Since 1 < r < 3, $r=\sqrt{2}\approx 1.414$ satisfies the given range.
Therefore, the value of r² is 2.