The equation of a circle C1C_1 is given by the equation x2+y2−4x−8y=0x^2 + y^2 - 4x - 8y = 0.

  1. (a)

    Find the coordinates of the center and radius of C1C_1.

  2. (b)

    Point A (0,8)A\ (0,8) lies on the circle C1C_1.

    Find the equation of tangent to the circle, C1C_1 at AA.

  3. (c)

    The circle, C1C_1 is reflected about the yy-axis.

    Find the equation that represents the reflected circle C2C_2.

  4. (d)

    A circle, C3C_3 has the equation (x−5)2+(y−8)2=5(x-5)^2 + (y-8)^2 = 5

    Explain if C1C_1 and C3C_3 intersect or do not intersect each other.

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Worked solution

9(a)

1. Group the $x$ and $y$ terms and complete the square for each

x2−4x+y2−8y=0x^2 - 4x + y^2 - 8y = 0
(x−2)2−22+(y−4)2−42=0(x - 2)^2 - 2^2 + (y - 4)^2 - 4^2 = 0
(x−2)2−4+(y−4)2−16=0(x - 2)^2 - 4 + (y - 4)^2 - 16 = 0
(x−2)2+(y−4)2=20(x - 2)^2 + (y - 4)^2 = 20

2. Identify the center and radius from the standard form $(x-a)^2 + (y-b)^2 = r^2$

Center (a,b)=(2,4)(a, b) = (2, 4)
Radius r=20=25r = \sqrt{20} = 2\sqrt{5}

Answer

Center: (2,4)(2, 4), Radius: 20\sqrt{20} or 252\sqrt{5}

9(b)

1. Calculate the gradient of the radius connecting the center to point $A$.

From part (a), the center of C1C_1 is (2,4)(2, 4).
Gradient of radius at A(0,8)=8−40−2=4−2=−2A(0, 8) = \frac{8 - 4}{0 - 2} = \frac{4}{-2} = -2

2. Determine the gradient of the tangent at $A$.

Since the tangent is perpendicular to the radius at the point of contact:
Gradient of tangent =−1−2=12= -\frac{1}{-2} = \frac{1}{2}

3. Form the equation of the tangent line.

Using the point-gradient form with A(0,8)A(0, 8):
y−8=12(x−0)y - 8 = \frac{1}{2}(x - 0)
y=12x+8y = \frac{1}{2}x + 8

Answer

y=12x+8y = \frac{1}{2}x + 8

9(c)

1. Find the center of the reflected circle $C_2$

From part (a), the center of C1C_1 is (2,4)(2, 4) and the radius squared is 2020.
When reflected about the yy-axis, the xx-coordinate of the center changes sign.
Center of C2=(−2,4)C_2 = (-2, 4)

2. Write the equation of the reflected circle $C_2$

The radius remains the same.
Equation of C2C_2: (x−(−2))2+(y−4)2=20(x - (-2))^2 + (y - 4)^2 = 20
(x+2)2+(y−4)2=20(x + 2)^2 + (y - 4)^2 = 20
Expanding this gives:
x2+4x+4+y2−8y+16=20x^2 + 4x + 4 + y^2 - 8y + 16 = 20
x2+y2+4x−8y=0x^2 + y^2 + 4x - 8y = 0

Answer

x2+y2+4x−8y=0x^2 + y^2 + 4x - 8y = 0 or (x+2)2+(y−4)2=20(x+2)^2 + (y-4)^2 = 20

9(d)

1. State the centre and radius of $C_1$ and $C_3$.

From part (a), for C1C_1:
Centre O1=(2,4)O_1 = (2, 4)
Radius r1=22+42=20=25r_1 = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}

For C3C_3:
Centre O3=(5,8)O_3 = (5, 8)
Radius r3=5r_3 = \sqrt{5}

2. Calculate the distance between the centres of the two circles.

Distance O1O3=(5−2)2+(8−4)2O_1O_3 = \sqrt{(5 - 2)^2 + (8 - 4)^2}
O1O3=32+42=25=5O_1O_3 = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

3. Calculate the sum and difference of the radii.

Sum of radii, r1+r3=25+5=35≈6.71r_1 + r_3 = 2\sqrt{5} + \sqrt{5} = 3\sqrt{5} \approx 6.71
Difference of radii, r1−r3=25−5=5≈2.24r_1 - r_3 = 2\sqrt{5} - \sqrt{5} = \sqrt{5} \approx 2.24

4. Compare the distance between centres with the sum and difference of radii to determine intersection.

Since r1−r3<O1O3<r1+r3r_1 - r_3 < O_1O_3 < r_1 + r_3 (i.e., 2.24<5<6.712.24 < 5 < 6.71), the circles intersect each other at two points.

Answer

Since the distance between the centres (55) is less than the sum of their radii (≈6.71\approx 6.71) and greater than the difference of their radii (≈2.24\approx 2.24), C1C_1 and C3C_3 intersect each other.

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