For what value ofawill 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? :(

Printable View

- Dec 18th 2010, 03:11 AMdannycCircle 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? :( - Dec 18th 2010, 03:16 AMProve It
Since it's a circle, the distance from the centre to any point will be the same.

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

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

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

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

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

$\displaystyle \displaystyle 10a + 10 = 40$

$\displaystyle \displaystyle a + 1 = 4$

$\displaystyle \displaystyle a= 3$. - Dec 18th 2010, 04:12 AMArchie Meade
The circle equation is Pythagoras' theorem applied to all points (x, y) on the circumference and the centre (2, -3).

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

Use the 2 given points to find "a"

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

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

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

therefore solve for

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

as "Prove It" showed.

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

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

Therefore

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

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

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

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

Quote:

And how would you find the radius too? :(

$\displaystyle (-1)^2+(a+3)^2=R^2\Rightarrow\ 1+6^2=R^2\Rightarrow\ R=\sqrt{37}$ - Dec 18th 2010, 04:43 AMdannyc
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 $\displaystyle (1 - 2)^2 + (3 + 3)^2 = r^2$ and so $\displaystyle r^2 = 37 $ (Or is this not legal since I'm using all 3 points at the same time?) - Dec 18th 2010, 04:54 AMArchie Meade
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....

$\displaystyle (x,y)$ 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). - Dec 18th 2010, 04:58 AMdannyc
Thanks Archie that makes sense. Also, I never seen the other equation form of the circle with "c" before, that's pretty clever too :)

- Dec 18th 2010, 05:05 AMArchie Meade
The other one is the "multiplied out" version of

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

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

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

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

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

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.