Get the complete, step-by-step math solution for: "If the area of the larger portion bounded between the curves x² + y² = 25 and y = |x - 1| is (1)/(4)(b π + c), b, c {N}, then b + c is equal to". Powered by SolveForX AI math tutor.
Step 2: Calculate intersection points
Substitute y=∣x−1∣ into the circle equation. This splits into two cases: y=x−1 for x≥1 and y=−(x−1)=1−x for x<1. Solving x2+(x−1)2=25 gives 2x2−2x−24=0, or x2−x−12=0, which factors to (x−4)(x+3)=0. Since x≥1, we take x=4. Then y=∣4−1∣=3. So, (4,3) is an intersection point. Solving x2+(1−x)2=25 gives 2x2−2x−24=0, or x2−x−12=0. Since x<1, we take x=−3. Then y=∣−3−1∣=4. So, (−3,4) is another intersection point.
Step 3: Calculate the area of the smaller portion
The area of the smaller portion is the area under the V-shaped curve between x=−3 and x=4, minus the area of the triangle formed by the origin and the intersection points, plus the area of the sector. Alternatively, it's the area of the sector formed by the origin and the intersection points, minus the area of the triangle formed by the origin and the intersection points, plus the area under the line y=∣x−1∣. The area of the sector formed by the origin and the points (−3,4) and (4,3) can be found using the angle θ between the vectors OA=(−3,4) and OB=(4,3). cosθ=5⋅5(−3)(4)+(4)(3)=25−12+12=0, so θ=2π. The area of the sector is 21r2θ=21(52)(2π)=425π. The area of the triangle formed by the origin and the points (−3,4) and (4,3) is 21∣x1y2−x2y1∣=21∣(−3)(3)−(4)(4)∣=21∣−9−16∣=225. The area under y=∣x−1∣ from x=−3 to x=4 is ∫−31(1−x)dx+∫14(x−1)dx=[x−2x2]−31+[2x2−x]14=(1−21)−(−3−29)+(216−4)−(21−1)=21+215+4+21=217+4=225. The area of the smaller region is the area of the sector minus the area of the triangle formed by the origin and the intersection points, plus the area under the V-shaped curve. This is Asmaller=425π−225+225=425π.
Step 6: Compare with the given form and find b+c
We are given that the area of the larger portion is 41(bπ+c). Comparing this with our calculated area 475π, we can see that bπ+c=75π. Therefore, b=75 and c=0. However, the problem states b,c∈N, which means c must be a natural number (positive integer). This implies that the interpretation of the smaller area calculation might need adjustment. Let's re-evaluate the area bounded by the curves. The area bounded by the curves is the area of the circle minus the area of the region defined by y=∣x−1∣ inside the circle. The area of the region bounded by the curves is the area of the sector AOB (where A=(−3,4) and B=(4,3)) plus the area of the triangle A(1,0)B plus the area of the triangle A(-3,4)C where C=(1,0) and B(4,3)D where D=(1,0). The area of the sector AOB is 425π. The area of the triangle A(1,0)B is 21base×height. The base is the distance between x=−3 and x=4 along the line y=0, which is 4−(−3)=7. The height is the y -coordinate of the vertex (1,0) relative to the line segment connecting (−3,4) and (4,3). This is not straightforward.
Let's reconsider the area bounded by the curves. The area of the region bounded by the curves is the area of the circle minus the area of the region defined by y=∣x−1∣ inside the circle. The area of the region defined by y=∣x−1∣ inside the circle is the area of the triangle formed by the points (−3,4), (1,0), and (4,3). This is a polygon. We can calculate its area using the shoelace formula or by dividing it into two triangles.
Area of triangle with vertices (−3,4), (1,0), (4,3):
Using shoelace formula: 21∣(−3)(0)+(1)(3)+(4)(4)−(4)(1)−(0)(4)−(3)(−3)∣ =21∣0+3+16−4−0−(−9)∣ =21∣19−4+9∣=21∣24∣=12.
This is the area of the region *under* the V-shape.
The area of the smaller portion is the area of the sector AOB minus the area of the triangle AOB.
Area of sector AOB=425π.
Area of triangle AOB=21∣(−3)(3)−(4)(4)∣=21∣−9−16∣=225.
So, the area of the smaller segment of the circle cut by the chord AB is 425π−225.
The area of the region bounded by the curves is the area of the triangle formed by the points (−3,4), (1,0), and (4,3) plus the area of the segment of the circle.
Area bounded by curves = 12+(425π−225)=425π+12−225=425π+224−25=425π−21.
This is the area of the region *inside* the circle and *above* the V-shape. This is the smaller portion.
Asmaller=425π−21.
Alarger=Acircle−Asmaller=25π−(425π−21)=25π−425π+21=475π+21.
We are given Alarger=41(bπ+c).
So, 475π+21=4bπ+4c.
Comparing coefficients:
4bπ=475π⟹b=75.
4c=21⟹c=2.
Both b=75 and c=2 are natural numbers.
Therefore, b+c=75+2=77.