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## Homework Statement

Find the centre of circle X2 + y2 + 4x – 16y + 18 = 0, show it's radius is 5√2 and find the co-ordinates where it is intersected by y = 3x - 6

## Homework Equations

## The Attempt at a Solution

X2 + y2 + 4x – 16y + 18 = 0

by completing the square using X2 + 2px = (x + p) 2 – p2

X^2 + y^2 + 4x – 16y + 18 = 0

Substituting p for 2 we get;

X^2 + 4x = (x + 2) ^2 – 2^2 = (x + 2) ^2 – 4

X^2 + y^2 + 4x – 16y + 18 = 0

Substituting p for -8 we get;

y^2 -16y = (y – 8) ^2 – (-8)^2 = (y – 8) ^2 – 64

Collecting the completed-square forms gives;

(x + 2) ^2 – 4 + (y – 8)^ 2 – 64 + 18 = 0

Collecting and rearranging gives;

(x + 2) ^2 + (y – 8) ^2 = 50

Comparing this to the form (x-a) ^2 + (y-b) ^2 = r^2 we can see that a = 2 and b = 8. The radius = √50 = 7.071067812.

√50 = 7.071067812.

√2 = 1.414213562 * 5 = 7.071067812.

To find the intersection I tried using substitution but have got myself confused.

(x + 2) ^2 + (y – 8) ^2 = 50, y = 3x - 6

(x + 2) ^2 + (3x – 14) ^2 = 50

4x^2 - 80x + 150 = 0

Now I can't work out how to factorise the above to gain an answer for X and so for Y also as co-ordinates.

I'm guessing that the line doesn't intersect the circle and but thought I should be able to take the calculation further to prove it, rather than relying on myself not being able to do it.

I would really appreciate any pointers as to where my mistake(s) might be. I am doing an adult ed course so please don't give me the answer.

thanks for reading.