The equation of a circle is x2+y2+2x−8y−8=0x^2 + y^2 + 2x - 8y - 8 = 0.

  1. (a)

    Find the coordinates of the centre, CC and the radius of the circle.

  2. (b)

    Find the equation of the tangent to the circle at the point P(−2,8)P(-2, 8).

  3. (c)

    If the circle is reflected in the line y=1y = 1, find the equation of the reflected circle.

  4. (d)

    Find the area of the biggest square that can be drawn inside the circle.

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

7(a)

1. Rearrange the equation by completing the square for $x$ and $y$.

x2+2x+y2−8y=8x^2 + 2x + y^2 - 8y = 8
(x+1)2−12+(y−4)2−(−4)2=8(x + 1)^2 - 1^2 + (y - 4)^2 - (-4)^2 = 8
(x+1)2−1+(y−4)2−16=8(x + 1)^2 - 1 + (y - 4)^2 - 16 = 8
(x+1)2+(y−4)2=25(x + 1)^2 + (y - 4)^2 = 25

2. Identify the coordinates of the centre.

Comparing with (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2,
Centre C=(−1,4)C = (-1, 4)

3. Identify the radius of the circle.

r2=25r^2 = 25
r=5r = 5

Answer

Centre C(−1,4)C(-1, 4), Radius =5= 5

7(c)

1. Find the centre and radius of the original circle from part (a)

From part (a), the original circle is x2+y2+2x−8y−8=0x^2 + y^2 + 2x - 8y - 8 = 0.
Completing the square:
(x+1)2−1+(y−4)2−16−8=0(x + 1)^2 - 1 + (y - 4)^2 - 16 - 8 = 0
(x+1)2+(y−4)2=25(x + 1)^2 + (y - 4)^2 = 25
Centre C=(−1,4)C = (-1, 4), Radius r=5r = 5.

2. Find the coordinates of the reflected centre

The line of reflection is y=1y = 1.
The xx-coordinate of the centre remains unchanged: x′=−1x' = -1.
The yy-coordinate of the centre is reflected across y=1y = 1. The distance from y=4y = 4 to y=1y = 1 is 33 units.
The reflected yy-coordinate is 1−3=−21 - 3 = -2.
New centre C′=(−1,−2)C' = (-1, -2).

3. Write the equation of the reflected circle

The radius remains the same, r=5r = 5.
The equation of the reflected circle is:
(x−(−1))2+(y−(−2))2=52(x - (-1))^2 + (y - (-2))^2 = 5^2
(x+1)2+(y+2)2=25(x + 1)^2 + (y + 2)^2 = 25
Expanding this gives:
x2+2x+1+y2+4y+4=25x^2 + 2x + 1 + y^2 + 4y + 4 = 25
x2+y2+2x+4y−20=0x^2 + y^2 + 2x + 4y - 20 = 0

Answer

(x+1)2+(y+2)2=25(x + 1)^2 + (y + 2)^2 = 25 or x2+y2+2x+4y−20=0x^2 + y^2 + 2x + 4y - 20 = 0

7(d)

1. Determine the diameter of the circle from its radius

From part (a), the radius of the circle is r=(−1)2+42−(−8)=5r = \sqrt{(-1)^2 + 4^2 - (-8)} = 5.
The diameter of the circle is d=2×5=10d = 2 \times 5 = 10.

2. Calculate the area of the inscribed square

The biggest square that can be drawn inside the circle is inscribed in it, meaning its diagonal is equal to the diameter of the circle.
Let the side length of the square be ss.
By Pythagoras' theorem, s2+s2=d2s^2 + s^2 = d^2
2s2=102=1002s^2 = 10^2 = 100
s2=50s^2 = 50
The area of the square is s2=50s^2 = 50.

Answer

5050 units2^2

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1The curve CC has equation y=x2+x+3x−2y = \frac{x^2 + x + 3}{x - 2}. Without using a calculator, find the set of values that yy…
  2. 2Find the gradient function of the curve y=2x2−ln⁡x+cy = 2x^2 - \ln x + c. The curve touches the positive xx-axis at point TT. Fi…
  3. 3The radius of a circle increases at a rate of 3 cm s^{-1}. Find, in terms of π\pi, the rate of increase of the area whe…
  4. 4定积分∫01(2x−x2)dx\int_{0}^{1}(\sqrt{2x{-}x^{2}})dx等于()({ }) A. π4\dfrac{\pi}{4} B. π2\dfrac{\pi}{2} C. π−14\dfrac{\pi{-}1}{4}…
  5. 5已知两定点M(0,1)M(0,1),N(1,2)N(1,2),平面内一动点PP到MM的距离与PP到NN的距离之比为2 \sqrt {2},直线y=kx−1y=kx-1与点PP的轨迹交于AA,BB两点. (…