1. ## Finding Circle

I'm having hard time finding equation of a circle with given two points and a radius.

given points (-2,3) and (2,-1) and radius = 4

my midpoint is
$(\frac{-2+2}{2},\frac{3-1}{2})$
=(0,1)

my distance is :

$d=\sqrt{(2+2)^2+(-1-3)^2}$
$=\sqrt{32}$

What should I do next to find the center and the equation of the circle?

2. Originally Posted by Anemori
I'm having hard time finding equation of a circle with given two points and a radius.

given points (-2,3) and (2,-1) and radius = 4

my midpoint is
$(\frac{-2+2}{2},\frac{3-1}{2})$
=(0,1)

my distance is :

$d=\sqrt{(2+2)^2+(-1-3)^2}$
$=\sqrt{32}$

What should I do next to find the center and the equation of the circle?

Let's assume that $k:\; (x-u)^2+(y-v)^2=16$ is the equation of the circle with center $C(u,v)$ and radius 4.

Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

$(-2-u)^2+(3-v)^2=16$ and $(2-u)^2+(-1-v)^2=16$.
If you now subtract the second equation from the first, you get $8u-8v+8=0$, and hence $u=v-1$.

Replacing $u$ in the first equation by $v-1$ gives you the quadratic equation $(-1-v)^2+(3-v)^2=16$.
Now solve for $v$: this gives you two solutions $v_1,v_2$ for $v$. Finally, pair these two solutions with the corresponding x-coordinates of C, namely $u_{1,2}=v_{1,2}-1$, and you are done.

3. Originally Posted by Failure
Let's assume that $k:\; (x-u)^2+(y-v)^2=16$ is the equation of the circle with center $C(u,v)$ and radius 4.

Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

$(-2-u)^2+(3-v)^2=16$ and $(2-u)^2+(-1-v)^2=16$.
If you now subtract the second equation from the first, you get $8u-8v+8=0$, and hence $u=v-1$.

Replacing $u$ in the first equation by $v-1$ gives you the quadratic equation $(-1-v)^2+(3-v)^2=16$.
Now solve for $v$: this gives you two solutions $v_1,v_2$ for $v$. Finally, pair these two solutions with the corresponding x-coordinates of C, namely $u_{1,2}=v_{1,2}-1$, and you are done.

im lost where u=v-1...

4. Originally Posted by Anemori
im lost where u=v-1...
Well, as I wrote, you can now replace u in $(-2-u)^2+(3-v)^2=16$ by v-1.

This gives you that $(-1-v)^2+(3-v)^2=16$ must hold. Thus it follows from this quadratic equation that $v_1=-1$ or $v_2=3$.

Since we know that $u=v-1$, we get for the corresponding x-coordinates $u_{1,2}$ of the center of the circle, that $u_1=v_1-1=-2$ and $u_2=v_2-1=2$, respectively.

Thus the center of the circle is either $C_1(-2,-1)$ or $C_2(2,3)$, and the corresponding equations of these two circles are

$k_1:\; (x+2)^2+(y+1)^2=4^2$ and $k_2:\; (x-2)^2+(y-3)^2=4^2$.