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The circle C has equation $$x^2 + y^2 - 10x + 6y + 30 = 0$$ Find (a) the coordinates of the centre of C, (b) the radius of C, (c) the y coordinates of the points where the circle C crosses the line with equation $x = 4$, giving your answers as simplified surds. - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 3

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The-circle-C-has-equation--$$x^2-+-y^2---10x-+-6y-+-30-=-0$$--Find--(a)-the-coordinates-of-the-centre-of-C,--(b)-the-radius-of-C,--(c)-the-y-coordinates-of-the-points-where-the-circle-C-crosses-the-line-with-equation-$x-=-4$,-giving-your-answers-as-simplified-surds.-Edexcel-A-Level Maths Pure-Question 7-2017-Paper 3.png

The circle C has equation $$x^2 + y^2 - 10x + 6y + 30 = 0$$ Find (a) the coordinates of the centre of C, (b) the radius of C, (c) the y coordinates of the point... show full transcript

Worked Solution & Example Answer:The circle C has equation $$x^2 + y^2 - 10x + 6y + 30 = 0$$ Find (a) the coordinates of the centre of C, (b) the radius of C, (c) the y coordinates of the points where the circle C crosses the line with equation $x = 4$, giving your answers as simplified surds. - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 3

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 start with the standard equation of a circle, which is in the form:

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

where (h,k)(h, k) is the center and rr is the radius.

  1. Rearranging the given equation:

    x2−10x+y2+6y+30=0x^2 - 10x + y^2 + 6y + 30 = 0

    Group the x and y terms:

    (x2−10x)+(y2+6y)=−30(x^2 - 10x) + (y^2 + 6y) = -30

  2. Complete the square for x:

    • Take half of -10 which is -5, square it to get 25.
    • Thus, we have:

    (x−5)2−25(x - 5)^2 - 25

  3. Complete the square for y:

    • Take half of 6 which is 3, square it to get 9.
    • Thus, we have:

    (y+3)2−9(y + 3)^2 - 9

  4. Substitute back:

    (x−5)2−25+(y+3)2−9=−30(x - 5)^2 - 25 + (y + 3)^2 - 9 = -30

  5. Simplifying this gives:

    (x−5)2+(y+3)2=4(x - 5)^2 + (y + 3)^2 = 4

The coordinates of the center are therefore (5, -3).

Step 2

the radius of C

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Answer

From the simplified equation of the circle:

(x−5)2+(y+3)2=4(x - 5)^2 + (y + 3)^2 = 4

It is clear that:

  • The radius squared, r2r^2, is equal to 4.

Thus, the radius rr is calculated as:

r=extsqrt(4)=2r = ext{sqrt}(4) = 2

Step 3

the y coordinates of the points where the circle C crosses the line x = 4

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Answer

To find where the circle crosses the line x=4x = 4, substitute x=4x = 4 in the circle's equation:

Starting with:

(x−5)2+(y+3)2=4(x - 5)^2 + (y + 3)^2 = 4

Substituting x=4x = 4 gives:

(4−5)2+(y+3)2=4(4 - 5)^2 + (y + 3)^2 = 4

Calculating:

1+(y+3)2=41 + (y + 3)^2 = 4

Subtracting 1:

(y+3)2=3(y + 3)^2 = 3

Now, taking the square root gives us two possible solutions:

y+3=extsqrt(3)extory+3=−extsqrt(3)y + 3 = ext{sqrt}(3) ext{ or } y + 3 = - ext{sqrt}(3)

From these:

  1. y=−3+extsqrt(3)y = -3 + ext{sqrt}(3)
  2. y=−3−extsqrt(3)y = -3 - ext{sqrt}(3)

Hence, the y-coordinates where the circle crosses the line x=4x = 4 are:

y=−3+extsqrt(3)extandy=−3−extsqrt(3)y = -3 + ext{sqrt}(3) ext{ and } y = -3 - ext{sqrt}(3)

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