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A circle has equation $x^2 + y^2 - 6x - 8y = 264$ - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 3

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A circle has equation $x^2 + y^2 - 6x - 8y = 264$. AB is a chord of the circle. The angle at the centre of the circle, subtended by AB, is 0.9 radians, as shown ... show full transcript

Worked Solution & Example Answer:A circle has equation $x^2 + y^2 - 6x - 8y = 264$ - AQA - A-Level Maths Pure - Question 5 - 2019 - Paper 3

Step 1

Find radius of the circle

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Answer

To find the radius of the circle, we start with the equation: x2+y26x8y=264x^2 + y^2 - 6x - 8y = 264 We can complete the square for the xx and yy terms:

  1. For x26xx^2 - 6x, we add and subtract (6/2)2=9(6/2)^2 = 9.
  2. For y28yy^2 - 8y, we add and subtract (8/2)2=16(8/2)^2 = 16.

Thus, we rewrite the equation as: (x3)2+(y4)2=289(x - 3)^2 + (y - 4)^2 = 289 This shows us that the center of the circle is at (3, 4) and the radius is:

r=extsqrt(289)=17r = ext{sqrt}(289) = 17.

Step 2

Find area of the sector

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Answer

The area of a sector (A) is given by: A=12r2θA = \frac{1}{2} r^2 \theta Substituting the radius (17) and angle (0.9 radians): A=12×172×0.9=130.05A = \frac{1}{2} \times 17^2 \times 0.9 = 130.05.

Step 3

Find area of triangle

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Answer

The area of triangle formed by the radius and chord (A_triangle) can be calculated using: Atriangle=12r2sin(θ)A_{triangle} = \frac{1}{2} r^2 \sin(\theta) Substituting the radius (17) and angle (0.9 radians): Atriangle=12×172×sin(0.9)113.19A_{triangle} = \frac{1}{2} \times 17^2 \times \sin(0.9) \approx 113.19.

Step 4

Find area of the minor segment

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Answer

The area of the minor segment (A_segment) is then given by: Asegment=AsectorAtriangleA_{segment} = A_{sector} - A_{triangle} Substituting the values calculated: Asegment=130.05113.19=16.86A_{segment} = 130.05 - 113.19 = 16.86 Rounded to three significant figures, the final answer is 16.9.

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