**14.** A circle \( C \) of radius 2 lies in the second quadrant and touches both the coordinate axes. Let \( r \) be the radius of a circle that has centre at the point \( (2, 5) \) and intersects the circle \( C \) at exactly two points. If the set of all possible values of \( r \) is the interval \( (\alpha, \beta) \), then \( 3\beta - 2\alpha \) is equal to: - (1) 15 - (2) 14 - (3) 12 - (4) 10

1. Placing the first circle (touching both axes)

When a circle touches the x-axis and y-axis, the distance from its centre to each axis is the radius.

• Because the circle lies in the second quadrant (x < 0, y > 0), its centre must be (−2, 2) (left of y-axis by 2, above x-axis by 2) and its radius is 2.

2. Distance between two fixed centres

The new circle’s centre is fixed at (2, 5). The straight-line distance d between the two centres is obtained by the distance formula

3. Condition for "exactly two" intersection points

For two circles with radii R and r (here R = 2) and centre-distance d = 5:

• They intersect in two points iff $|r - R| < d < r + R.$ (Strict inequalities because touching in one point gives equality.)

• Plug in R = 2 and d = 5 to get two separate inequalities:

  1. $|r - 2| < 5$
  2. $5 < r + 2$

4. Solving the inequalities (the student’s chain of logic)

1. First inequality: $|r - 2| < 5$ means
   $-5 < r - 2 < 5 \;\Longrightarrow\; -3 < r < 7.$
   (Since radius can’t be negative, we will throw away negative options later.)
2. Second inequality: $5 < r + 2 \;\Longrightarrow\; r > 3.$
3. Combine the two results and the fact r > 0: $r > 3 \quad\text{and}\quad r < 7 \;\Longrightarrow\; r \in (3, 7).$

Hence α = 3, β = 7 for the interval (α, β) = (3, 7).

Finally, the value demanded is
$3β - 2α = 3(7) - 2(3) = 21 - 6 = 15.$

Imagine two soap bubbles on graph paper

1. Big picture: One bubble is already drawn in the top-left part (second quadrant). It just touches both the edges of the paper (the x-axis and y-axis) and has a fixed size (radius 2).
2. New bubble: We want to blow another bubble with its centre at (2, 5). We can decide how big this bubble is (that’s the radius r).
3. Goal: We must make the new bubble overlap the old one in exactly two points (so they cross, but don’t just kiss or stay apart).
4. What we’re hunting: All the possible sizes r that work form a stretch of numbers—an interval (α, β). The question finally asks us to plug α and β into 3β − 2α.

So the job is mainly: "When do two bubbles cross in exactly two spots?" Then do a little number-crunching.

Step-by-Step Solution

1. First circle: Centre at $(−2, 2)$ with radius $2$ (because it touches both axes in the second quadrant).

2. Centre separation:

   $$d = \sqrt{(2 - (−2))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = 5.$$

3. Intersection condition for exactly two points:

   $|r - 2| < 5 < r + 2.$

4. Solve the left part:

   $|r - 2| < 5 \Rightarrow -5 < r - 2 < 5 \Rightarrow -3 < r < 7.$

5. Solve the right part:

   $5 < r + 2 \Rightarrow r > 3.$

6. Combine and keep $r > 0$:

   $r \in (3, 7).$

   So $\alpha = 3$, $\beta = 7$.

7. Compute required expression:

   $3\beta - 2\alpha = 3(7) - 2(3) = 21 - 6 = 15.$

Final answer: 15 (Option 1).