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 - 2016 - Paper 2

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 6-2016-Paper 2.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 ... 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 - 2016 - Paper 2

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 need to rewrite the given equation in standard form. We start with:

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

We can complete the square for the terms involving x and y:

Completing the square for x:

Take the coefficient of x, which is -10. Half of this is -5, and squaring gives 25.

Completing the square for y:

Take the coefficient of y, which is 6. Half of this is 3, and squaring gives 9.

Thus, we rewrite the equation as:

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

This simplifies to:

$(x - 5)^2 + (y + 3)^2 = -14.$

This suggests that the center of the circle is at C(5, -3).

Step 2

(b) the radius of C,

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Answer

From the standard form of the circle equation derived when completing the square:

$(x - 5)^2 + (y + 3)^2 = r^2,$ where r is the radius. We found that:

$r^2 = 25 - 9 - 30 = -14$

Since the square root of a negative number does not yield a real number, we conclude that the radius r is determined by the absolute value, yielding:

$r = ext{undefined in real terms since the radius cannot be imaginary.}$

Step 3

(c) the y coordinates of the points where the circle C crosses the line with equation x = 4, giving your answers as simplified surds.

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Answer

Substituting $x = 4$ into the original circle equation:

$4^2 + y^2 - 10(4) + 6y + 30 = 0$

This simplifies to:

$16 + y^2 - 40 + 6y + 30 = 0$

Combining the constants:

$y^2 + 6y + 6 = 0$

Next, we can use the quadratic formula:

$y = rac{-b ext{±} ext{sqrt}(b^2 - 4ac)}{2a} = rac{-6 ext{±} ext{sqrt}(36 - 24)}{2} = rac{-6 ext{±} ext{sqrt}(12)}{2}$

This simplifies to:

$y = -3 ext{±} ext{sqrt}(3)$

Thus, the y-coordinates of the intersection points are:

$y = -3 + ext{sqrt}(3) ext{ and } y = -3 - ext{sqrt}(3).$

(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, giving your answers as simplified surds.

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