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18 (a) Describe fully the graph of $x^2 + y^2 = 20$ - OCR - GCSE Maths - Question 18 - 2023 - Paper 6

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18 (a) Describe fully the graph of $x^2 + y^2 = 20$. (b) The graph of $y = 3x + 10$ intersects the graph of $x^2 + y^2 = 20$ at two points. Use an algebra... show full transcript

Worked Solution & Example Answer:18 (a) Describe fully the graph of $x^2 + y^2 = 20$ - OCR - GCSE Maths - Question 18 - 2023 - Paper 6

Step 1

Describe fully the graph of $x^2 + y^2 = 20$

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Answer

The equation x2+y2=20x^2 + y^2 = 20 represents a circle centered at the origin (0, 0) with a radius of ext{r} = rac{ ext{d}}{2} = rac{ ext{area}}{20}. Therefore, the radius is:

\Rightarrow r \approx 4.47 $$ The circle includes all points (x, y) that are at a distance of approximately 4.47 units from the center of the circle.

Step 2

Use an algebraic method to work out the coordinates of the two points

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Answer

To find the points of intersection of the equations y=3x+10y = 3x + 10 and x2+y2=20x^2 + y^2 = 20, we substitute yy from the linear equation into the circular equation:

  1. Substitute: x2+(3x+10)2=20x^2 + (3x + 10)^2 = 20

    This expands to: x2+(9x2+60x+100)=20x^2 + (9x^2 + 60x + 100) = 20

    Giving us: 10x2+60x+100−20=010x^2 + 60x + 100 - 20 = 0

    Simplifying: 10x2+60x+80=010x^2 + 60x + 80 = 0

    Dividing the entire equation by 10 yields: x2+6x+8=0x^2 + 6x + 8 = 0

  2. Factor the quadratic: (x+2)(x+4)=0(x + 2)(x + 4) = 0

  3. Solve for x: x=−2, x=−4x = -2, \, x = -4

  4. Substitute back to find y coordinates:

    For x=−2x = -2: y=3(−2)+10=4y = 3(-2) + 10 = 4

    For x=−4x = -4: y=3(−4)+10=−2y = 3(-4) + 10 = -2

Thus, the points of intersection are (-2, 4) and (-4, -2).

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