A-Level Edexcel Maths Pure Numerical Methods: The circle C has equation $x^2

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 6 - 2017 - Paper 3

Question 6

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 6-2017-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 points where... 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 6 - 2017 - Paper 3

Step 1

a) the coordinates of the centre of C

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Answer

To find the coordinates of the center of circle C, we must rewrite the equation of the circle into standard form. The standard form of a circle's equation is given by:

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

where ((h, k)) represents the center and (r) the radius.

Starting with the equation:

$x^2 + y^2 - 10x + 6y + 30 = 0$

We can rearrange and complete the square:

Rearrange terms involving (x) and (y): $x^2 - 10x + y^2 + 6y = -30$
Complete the square for (x): $x^2 - 10x = (x - 5)^2 - 25$
Complete the square for (y): $y^2 + 6y = (y + 3)^2 - 9$
Substitute back into the equation: $(x - 5)^2 - 25 + (y + 3)^2 - 9 = -30$ This simplifies to: $(x - 5)^2 + (y + 3)^2 = 4$

From this, the coordinates of the center are:

Center: (5, -3)

Step 2

b) the radius of C

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Answer

From the equation obtained in part (a):

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

we can see that the radius (r) can be found since $r^2 = 4$ . Thus, the radius is:

$r = ext{sqrt{4}} = 2$

Radius: 2

Step 3

c) the y coordinates of the points where the circle C crosses the line with equation x = 4

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Answer

To find the y-coordinates where the circle crosses the line $x = 4$ , substitute (x = 4) into the circle equation:

From the equation:

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

Substituting (x = 4):

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

This simplifies to:

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

Which further simplifies to:

$(y + 3)^2 = 4 - 1$

$(y + 3)^2 = 3$

Taking the square root of both sides gives:

$y + 3 = ext{±} ext{sqrt{3}}$

Solving for (y):

$y = -3 + ext{sqrt{3}}$
$y = -3 - ext{sqrt{3}}$

Thus, the y-coordinates where the circle crosses the line are:

Answers: (-3 + ext{sqrt{3}}, -3 - ext{sqrt{3}}

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

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