Circle problem

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December 18th 2010, 03:11 AM
dannyc

Circle problem

For what value of a will a circle centered at (2, -3) pass through points (1,a) and (a,3) in the standard (x,y) plane?

The answer should be 3, but how?? And how would you find the radius too? :(
December 18th 2010, 03:16 AM
Prove It

Since it's a circle, the distance from the centre to any point will be the same.

So $\displaystyle \sqrt{(2-1)^2 + (-3-a)^2} = \sqrt{(2-a)^2 + (-3-3)^2}$

$\displaystyle \sqrt{1 + 9 + 6a + a^2} = \sqrt{4 - 4a + a^2 + 36}$

$\displaystyle \sqrt{a^2 + 6a + 10} = \sqrt{a^2 - 4a + 40}$

$\displaystyle a^2 + 6a + 10 = a^2 - 4a + 40$

$\displaystyle 6a + 10 = -4a + 40$

$\displaystyle 10a + 10 = 40$

$\displaystyle a + 1 = 4$

$\displaystyle a= 3$ .
December 18th 2010, 04:12 AM
Archie Meade

Quote:

Originally Posted by dannyc

For what value of a will a circle centered at (2, -3) pass through points (1,a) and (a,3) in the standard (x,y) plane?

The answer should be 3, but how??

The circle equation is Pythagoras' theorem applied to all points (x, y) on the circumference and the centre (2, -3).

$\left(x-x_c\right)^2+\left(y-y_c\right)^2=R^2$

Use the 2 given points to find "a"

$(x-2)^2+(y--3)^2=R^2$

$(1-2)^2+(a+3)^2=R^2\;\;for\;\;(x,y)=(1,a)$

$(a-2)^2+(3+3)^2=R^2\;\;for\;\;(x,y)=(a,3)$

therefore solve for

$(a-2)^2+6^2=(-1)^2+(a+3)^2$

as "Prove It" showed.

Alternatively, if you use the $x^2+y^2+2gx+2fy+c=0$ form

then the centre is $(-g,-f)=(2,-3)$

Therefore

$1+a^2-4(1)+6a+c=0$

$a^2+9-4a+6(3)+c=0$

In solving these, $a^2$ and $c$ are eliminated, to give

$6a-3=27-4a\Rightarrow\ 10a=30$

Quote:

And how would you find the radius too? :(

Given "a", you can then calculate the radius

$(-1)^2+(a+3)^2=R^2\Rightarrow\ 1+6^2=R^2\Rightarrow\ R=\sqrt{37}$
December 18th 2010, 04:43 AM
dannyc

Thanks you guys! :) I get it now :) Just a quick thought though, without knowing what a is, could you have found the radius by taking the x coordinate from both points and taking the y coordinate from both points?
I still get $(1 - 2)^2 + (3 + 3)^2 = r^2$ and so $r^2 = 37$ (Or is this not legal since I'm using all 3 points at the same time?)
December 18th 2010, 04:54 AM
Archie Meade

Quote:

Originally Posted by dannyc

Thanks you guys! :) I get it now :) Just a quick thought though, without knowing what a is, could you have found the radius by taking the x coordinate from both points and taking the y coordinate from both points?
I still get $(1 - 2)^2 + (3 + 3)^2 = r^2$ and so $r^2 = 37$ (Or is this not legal since I'm using all 3 points at the same time?)

No, it doesn't work,
but you could get marked correctly in a test if the teacher did not check your work!
or if you did not state that that is what you were doing!

It would probably have been assumed that since you found the value of "a"
you were now using the point (1,a)=(1,3)

but to use y from one point and x from another point is incorrect....

$(x,y)$ refers to the horizontal and vertical co-ordinates of any "specific"
point on the circumference of the circle.

For example, if a circle circumference contains the points (4,3) and (0,5)
then it doesn't follow that it contains the point (4,5).
December 18th 2010, 04:58 AM
dannyc

Thanks Archie that makes sense. Also, I never seen the other equation form of the circle with "c" before, that's pretty clever too :)
December 18th 2010, 05:05 AM
Archie Meade

Quote:

Originally Posted by dannyc

Thanks Archie that makes sense. Also, I never seen the other equation form of the circle with "c" before, that's pretty clever too :)

The other one is the "multiplied out" version of

$(x+g)^2+(y+f)^2=R^2$

where the centre co-ordinates are labelled $(-g,-f)$ instead of $\left(x_c,y_c\right)$

Then, both sides being equal, they are subtracted...

$(x+g)^2+(y+f)^2-R^2=0$

so the "c" will be $g^2+f^2-R^2$

That version is often employed to find centre co-ordinates and radius
by algebraic solving of simultaneous equations.
It's just a convention though. No different to the first method.