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:

Step 1: Determine the center of circle C

Since circle $C$ has a radius of 2 and lies in the second quadrant, touching both the $x$ -axis and $y$ -axis, its center must be at $(-2, 2)$. The second quadrant is where $x$ coordinates are negative and $y$ coordinates are positive.

Step 2: Calculate the distance between the centers

Let $C_1 = (-2, 2)$ be the center of the first circle and $C_2 = (2, 5)$ be the center of the second circle. We use the distance formula to find the distance $d$ between these two centers.

Step 3: Compute the distance d

Substituting the coordinates into the distance formula, we get $d = \sqrt{(2 - (-2))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

Step 4: Determine the range of r for two intersection points

For two circles to intersect at exactly two points, the distance between their centers ($d$) must be greater than the absolute difference of their radii and less than the sum of their radii. Here, $r_1 = r$ and $r_2 = 2$.

Step 5: Solve the inequalities for r

We solve the two inequalities separately. First, $|r - 2| < 5$ implies $-5 < r - 2 < 5$, which simplifies to $-3 < r < 7$. Second, $5 < r + 2$ implies $r > 3$. Combining these conditions, we get $3 < r < 7$.

Step 6: Identify alpha and beta and calculate the final expression

From the interval $(3, 7)$, we have $\alpha = 3$ and $\beta = 7$. Now we substitute these values into the expression $3\beta - 2\alpha$.

Step 7: Final Calculation

Performing the final calculation, $3(7) - 2(3) = 21 - 6 = 15$.

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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