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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 show... 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 the Radius of the Circle

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Answer

To find the radius of the circle, we start with the standard form of the circle's equation. First, complete the square for the equation:

x26x+y28y=264x^2 - 6x + y^2 - 8y = 264

Completing the square:

  1. For x26xx^2 - 6x:

    • Take half of -6, square it: igg(-\frac{6}{2}\bigg)^2 = 9
    • Thus, x26xx^2 - 6x becomes (x3)29(x - 3)^2 - 9.
  2. For y28yy^2 - 8y:

    • Take half of -8, square it: igg(-\frac{8}{2}\bigg)^2 = 16
    • Thus, y28yy^2 - 8y becomes (y4)216(y - 4)^2 - 16.

Putting this back into the equation, we have:

(x3)29+(y4)216=264(x - 3)^2 - 9 + (y - 4)^2 - 16 = 264

So,

(x3)2+(y4)2=289(x - 3)^2 + (y - 4)^2 = 289

The radius is: r=289=17r = \sqrt{289} = 17

Step 2

Find the Area of the Sector

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Answer

The area of the sector can be found using the formula: Area of Sector=12r2θ\text{Area of Sector} = \frac{1}{2} r^2 \theta where r=17r = 17 and θ=0.9\theta = 0.9 radians:

Area of Sector=12×172×0.9=12×289×0.9=130.05\text{Area of Sector} = \frac{1}{2} \times 17^2 \times 0.9 = \frac{1}{2} \times 289 \times 0.9 = 130.05

Step 3

Find the Area of the Triangle

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Answer

The area of triangle OABOAB can be found using: Area=12r2sin(θ)\text{Area} = \frac{1}{2} r^2 \sin(\theta) where r=17r = 17 and θ=0.9\theta = 0.9 radians:

Area=12×172×sin(0.9)=12×289×0.621689.86\text{Area} = \frac{1}{2} \times 17^2 \times \sin(0.9) = \frac{1}{2} \times 289 \times 0.6216 \approx 89.86

Step 4

Calculate the Area of the Minor Segment

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Answer

Now, we can find the area of the minor segment by subtracting the area of the triangle from the area of the sector:

Area of Minor Segment=Area of SectorArea of Triangle\text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle}

Substituting the values: Area of Minor Segment=130.0589.8640.19\text{Area of Minor Segment} = 130.05 - 89.86 \approx 40.19

Finally, rounding to three significant figures, the area is approximately 40.2.

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