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The circle C has equation $x^2 + y^2 + 4x - 2y - 11 = 0$ Find (a) the coordinates of the centre of C, (b) the radius of C, (c) the coordinates of the points where C crosses the y-axis, giving your answers as simplified surds. - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 2

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The-circle-C-has-equation-$x^2-+-y^2-+-4x---2y---11-=-0$---Find---(a)-the-coordinates-of-the-centre-of-C,---(b)-the-radius-of-C,---(c)-the-coordinates-of-the-points-where-C-crosses-the-y-axis,-giving-your-answers-as-simplified-surds.-Edexcel-A-Level Maths Pure-Question 6-2011-Paper 2.png

The circle C has equation $x^2 + y^2 + 4x - 2y - 11 = 0$ Find (a) the coordinates of the centre of C, (b) the radius of C, (c) the coordinates of the points ... show full transcript

Worked Solution & Example Answer:The circle C has equation $x^2 + y^2 + 4x - 2y - 11 = 0$ Find (a) the coordinates of the centre of C, (b) the radius of C, (c) the coordinates of the points where C crosses the y-axis, giving your answers as simplified surds. - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 2

Step 1

the coordinates of the centre of C

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Answer

To find the coordinates of the center of the circle, we first rewrite the equation in the standard form. Starting from:

x2+y2+4x2y11=0x^2 + y^2 + 4x - 2y - 11 = 0

we group the x and y terms:

(x2+4x)+(y22y)=11 (x^2 + 4x) + (y^2 - 2y) = 11

Next, we complete the square for the x terms and the y terms. For the x terms:

x2+4x=(x+2)24x^2 + 4x = (x + 2)^2 - 4

For the y terms:

y22y=(y1)21y^2 - 2y = (y - 1)^2 - 1

Putting these back into the equation gives us:

\Rightarrow (x + 2)^2 + (y - 1)^2 = 16 $$ From this standard form, the center of the circle C is at the coordinates (-2, 1).

Step 2

the radius of C

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Answer

From the standard form of the circle:

(x+2)2+(y1)2=16(x + 2)^2 + (y - 1)^2 = 16

we see that the radius, r, is the square root of 16:

r=extsqrt(16)=4r = ext{sqrt}(16) = 4

Thus, the radius of circle C is 4.

Step 3

the coordinates of the points where C crosses the y-axis

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Answer

To find where the circle crosses the y-axis, we set x = 0 in the equation.

Substituting x = 0 into the rearranged circle equation:

\Rightarrow 4 + (y - 1)^2 = 16 \ \Rightarrow (y - 1)^2 = 12 \ \Rightarrow y - 1 = ext{±} ext{sqrt}(12) \ \Rightarrow y = 1 ext{±} 2 ext{sqrt}(3) $$ Thus, the coordinates where the circle crosses the y-axis are: $$ (0, 1 + 2 ext{sqrt}(3)) \ (0, 1 - 2 ext{sqrt}(3)) $$

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