Question

help please....
anyone know how to work this out:

The circle C, with centre at the point A, has equation x2 + y2 – 10x + 9 = 0.

Find

(a) the coordinates of A,

-
i tried subsituting in x=0 but it didnt work?

Answer 1

** \[x^2+y^2-10x+9=0\]

Answer 2

\[(x-a)^2+(y-b)^2=r^2\]

\[x^2-10x+y^2+9=0\]\[x^2-10x+y^2+9+16-16=0\]\[x^2-10x+y^2+25=4^2\]\[(x-5)^2+(y-0)^2=4^2\]
here we can see the coordinates for A in the y direction will be 0 and for the x direction will be 5

So A is at the point (5,0)

Answer 3

where did you get the 16 from?

Answer 4

we have to complete the square to get it in the right form

Answer 5

@kinzan123,

That method is called "Completing the Square".

Answer 6

thanks :) ...with completing the square, you start by factorising right?

Answer 7

yes but it's of a special form (x-a)^2 that's why you complete the square

Answer 8

yeh.. so you get (x+4)^2 right?

Answer 9

then you have to takewaway the 16 to make it equal?

Answer 10

close so we had x^2-10x+9 let's ignore the y for now,,

x^2-10x+9+16-16
=x^2-10x+25-16
=(x-5)^2-16

because (x-a)^2=x^2-2ax+a^2

Answer 11

No, Completing the Square actually completes the square (just as what it is called).
Factorization comes at a later point when you have completed the square.

The process involves a binomial of a variable plus a constant (not always permanently).

Check this example.
If you have \[(x+4)^{2}\]
That would then be equal to
\[x^2+8x+16\]

Reversing the process would be Completing the Square.
Taking for example
\[x^2−10x+y^2+9=0\]
You add 16 in order to make 9 be 25, and then subtract 16 in order to balance the equation
\[x^2−10x+y^2+9+16−16=0\]
At this point, you can then have 9+16 as 25 and transpose (sorry about this term) 16 to the right side of the equation and simplify it to its square root squared.
\[x^2−10x+y^2+25=4^2\]

Then group the equation
\[x^2−10x+25+y^2=4^2\]
Then factorization comes along:
\[(x−5)^2+y^2=4^2\]

Then just simplify further and you have it.
Hope you understood.

Answer 12

@sercczionelabus
but why would you want to make 25? is it because you factoeised the x terms to get (x-5)^2

Answer 13

Because of the middle term that is -10.
If you have
\[ax^2+bx+c\]
Use this to get the constant that would make the quadratic equation a perfect squared binomial
\[(b/2)^2\]

Let's take a look back at the previous equation.
\[x^2−10x+y^2+9=0\]

Ignoring y and using the formula I gave you, you would have
\[(-10/2)^2=25\]

So you must have a constant of magnitude 25 in order to complete the square, and since you've got 9 already; all you have to do then is add 16.

Answer 14

ohhh , so in general ..do i always use (b/2)^2 ??

Answer 15

Yea. It is a very useful formula.
Though it's not a standard formula, it is just a formula based on analysis.

Answer 16

oh okay, thank you very much :)

Answer 17

Sure, no problem.

Answer 18

how would we simplify the following further?:
\[(x-5)^2+y^2=4^2\]

Answer 19

That is actually very much simplified.
But since you're dealing with a circle, you have to take note of the general form,
\[(x-h)^2+(y-k)^2\]
(Not sure if I have interchanged k and h).

So you can then add 0 to the left side of the equation (wouldn't matter since it is a zero and would still render the equation balanced).

And\[y^2+0\] is just the same as\[(y+0)^2\]

Upon writing it that way, you have now formulated it in such a way that it matches the general form. Then from h and k, you can then take the coordinates of your A (I guess, not sure about this since I forgot about this).

Answer 20

I forgot to put =r^2 at the last part of the general form. Sorry, my bad.

Answer 21

lol thanks :)

Answer 22

Sure.