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A) To determine the coordinates of A and B, we substitute the equation of the straight line y = x + 2 into the equation of the circle x2 + y2 = 20.

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6 parts
a

A(−4,−2) and B(2,4)\text{A}(-4, -2) \text{ and B}(2, 4)

b

62 units6\sqrt{2} \text{ units}

c
(-1, 1)
d

OM ⊥ AB\text{OM } \perp \text{ AB}

e

2x+y=−10 and x+2y=102x + y = -10 \text{ and } x + 2y = 10

f
(-10, 10)
Graph of y = x + 2

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Key points of y = x + 2
Pointxy
x-intercept−20
y-intercept02

Step-by-step solution

14 steps

Part a· 3 steps

To determine the coordinates of A and B, we substitute the equation of the straight line y=x+2y = x + 2 into the equation of the circle x2+y2=20x^2 + y^2 = 20.

  1. Substitute y=x+2y = x + 2 into the circle equation.

    x2+(x+2)2=20x^2 + (x + 2)^2 = 20 x2+(x2+4x+4)=20x^2 + (x^2 + 4x + 4) = 20 2x2+4x+4−20=02x^2 + 4x + 4 - 20 = 0 2x2+4x−16=02x^2 + 4x - 16 = 0

    Divide by 2:

    x2+2x−8=0x^2 + 2x - 8 = 0

  2. Factor the quadratic equation to find the x-coordinates.

    (x+4)(x−2)=0(x + 4)(x - 2) = 0

    So, x=−4x = -4 or x=2x = 2.

  3. Substitute the x-values back into y=x+2y = x + 2 to find the corresponding y-coordinates.

    For x=−4x = -4: y=−4+2=−2y = -4 + 2 = -2. So, point A is (−4,−2)(-4, -2).
    For x=2x = 2: y=2+2=4y = 2 + 2 = 4. So, point B is (2,4)(2, 4).
    The coordinates of A and B are A(−4,−2) and B(2,4)\boxed{\text{A}(-4, -2) \text{ and B}(2, 4)}.

Part b

62 units6\sqrt{2} \text{ units}

To determine the length of the chord AB, we use the distance formula between points A and B.

  1. Apply the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} for A(−4,−2)(-4, -2) and B(2,4)(2, 4).

    AB=(2−(−4))2+(4−(−2))2AB = \sqrt{(2 - (-4))^2 + (4 - (-2))^2} AB=(6)2+(6)2AB = \sqrt{(6)^2 + (6)^2} AB=36+36AB = \sqrt{36 + 36} AB=72AB = \sqrt{72}

  2. Simplify the square root.

    AB=36×2AB = \sqrt{36 \times 2} AB=62AB = 6\sqrt{2}

    The length of the chord AB is 62 units\boxed{6\sqrt{2} \text{ units}}.

Part c(-1, 1)

To determine the coordinates of M, the midpoint of the chord AB, we use the midpoint formula.

  1. Apply the midpoint formula M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) for A(−4,−2)(-4, -2) and B(2,4)(2, 4).

    M=(−4+22,−2+42)M = \left(\frac{-4 + 2}{2}, \frac{-2 + 4}{2}\right) M=(−22,22)M = \left(\frac{-2}{2}, \frac{2}{2}\right) M=(−1,1)M = (-1, 1)

    The coordinates of M are (−1,1)\boxed{(-1, 1)}.

Part d

OM ⊥ AB\text{OM } \perp \text{ AB}

To show that OM ⊥\perp AB, we need to check if the product of their slopes is -1. O is the origin (0,0)(0, 0).

  1. Calculate the slope of OM.

    mOM=yM−yOxM−xO=1−0−1−0=1−1=−1m_{OM} = \frac{y_M - y_O}{x_M - x_O} = \frac{1 - 0}{-1 - 0} = \frac{1}{-1} = -1

  2. Calculate the slope of AB.

    The line AB is y=x+2y = x + 2, so its slope is 1.

    mAB=1m_{AB} = 1

  3. Multiply the slopes.

    mOM×mAB=(−1)×(1)=−1m_{OM} \times m_{AB} = (-1) \times (1) = -1

    Since the product of the slopes is −1-1, OM is perpendicular to AB. OM ⊥ AB\boxed{\text{OM } \perp \text{ AB}}.

Part e

2x+y=−10 and x+2y=102x + y = -10 \text{ and } x + 2y = 10

To determine the equations of the tangents to the circle x2+y2=20x^2 + y^2 = 20 at A and B, we use the tangent formula xx1+yy1=r2x x_1 + y y_1 = r^2. Here, r2=20r^2 = 20.

  1. Find the equation of the tangent at A(−4,−2)(-4, -2).

    x(−4)+y(−2)=20x(-4) + y(-2) = 20 −4x−2y=20-4x - 2y = 20

    Divide by -2:

    2x+y=−102x + y = -10

  2. Find the equation of the tangent at B$(2, 4)$.

    x(2)+y(4)=20x(2) + y(4) = 20 2x+4y=202x + 4y = 20

    Divide by 2:

    x+2y=10x + 2y = 10

    The equations of the tangents are 2x+y=−10 and x+2y=10\boxed{2x + y = -10 \text{ and } x + 2y = 10}.

Part f(-10, 10)

To determine the coordinates of C, the point of intersection of the tangents, we solve the system of equations from part (e).

  1. Set up the system of equations.

    1. 2x+y=−102x + y = -10
    2. x+2y=10x + 2y = 10

  2. Solve for x and y.

    From equation (1), y=−2x−10y = -2x - 10. Substitute this into equation (2).

    x+2(−2x−10)=10x + 2(-2x - 10) = 10 x−4x−20=10x - 4x - 20 = 10 −3x=30-3x = 30 x=−10x = -10

  3. Substitute x=−10x = -10 back into y=−2x−10y = -2x - 10.

    y=−2(−10)−10y = -2(-10) - 10 y=20−10y = 20 - 10 y=10y = 10

    The coordinates of C are (−10,10)\boxed{(-10, 10)}.

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