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 (α,β), then 3β−2α is equal to:
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Step-by-Step Solution
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 C1=(−2,2) be the center of the first circle and C2=(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=(2−(−2))2+(5−2)2=42+32=16+9=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, r1=r and r2=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 α=3 and β=7. Now we substitute these values into the expression 3β−2α.
Step 7: Final Calculation
Performing the final calculation, 3(7)−2(3)=21−6=15.