Finding the Centre and Intersection of Circle X2 + y2 + 4x – 16y + 18 = 0

[Rubydeemer]

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.

 

[Pinu7]

Two things:
1. You should have been able to show that
$$\sqrt {50} = 5\sqrt 2$$ without a calculator. Try to do prime factorization to get
$$\sqrt {50} = \sqrt {5 \cdot 5 \cdot 2} = \sqrt {{5^2} \cdot 2} = 5\sqrt 2$$


2. On your last line, it is not supposed to be 2x^2, check your arithmetic. Also, just because you do not know how to factor it, doesn't mean there is no solution. You may have to use the quadratic formula. Of course, there could be both real and imaginary roots(for a total of two).

Cases:
1. Two real roots: the line would intersect the circle at two points
2. One real root, one imaginary root: the line would intersect the circle at one point(it would be tangent to the circle)
3. Two imaginary roots: the line will NOT intersect the circle.

 

[Rubydeemer]

Thanks Pinu7, I had tried it with quadratic formula and got x = 1734 and x = 1254 so thought I was way off.
I'll try again.

On your other point "On your last line, it is not supposed to be 2x^2, check your arithmetic" did you mean 4x^2?

thanks again.

 

[Dick]

Thanks Pinu7, I had tried it with quadratic formula and got x = 1734 and x = 1254 so thought I was way off.
I'll try again.

On your other point "On your last line, it is not supposed to be 2x^2, check your arithmetic" did you mean 4x^2?

thanks again.

Yes, it's not supposed to be 4x^2. What is it supposed to be? Be careful expanding (3x-14)^2. If you get it right, it factors easily.