A-Level Edexcel Maths Pure Laws of Logarithms: 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 7 - 2016 - Paper 2

Question 7

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 7-2016-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 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 - 2016 - Paper 2

Step 1

(a) the coordinates of the centre of C,

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Answer

To find the center of the circle, we need to rewrite the equation in standard form:

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

We complete the square for both the x and y terms. Starting with the x terms:

$x^2 - 10x = (x - 5)^2 - 25$

And for the y terms:

$y^2 + 6y = (y + 3)^2 - 9$

Substituting these back into the equation gives us:

$(x - 5)^2 - 25 + (y + 3)^2 - 9 + 30 = 0$

Simplifying this, we have:

$(x - 5)^2 + (y + 3)^2 - 4 = 0$

Therefore, we get:

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

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

Step 2

(b) the radius of C,

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Answer

From the standard form of the circle's equation we derived:

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

The radius is the square root of 4. Thus,

$r = ext{sqrt}(4) = 2.$

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 intersects with the line $x = 4$ , we substitute $x = 4$ into the circle's equation:

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

This simplifies to:

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

Rearranging gives:

$(y + 3)^2 = 3$

Taking the square root of both sides yields:

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

Thus, solving for y gives:

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

(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,

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