1. ## equation of circle

Find the equation of the circle with center at (-2, 3) and which passes through (-1, 4).

$(x+1)^2 + (y-3)^2=1$

Hows it look?

2. Originally Posted by R3ap3r
Find the equation of the circle with center at (-2, 3) and which passes through (-1, 4).

$(x+1)^2 + (y-3)^2=1$

Hows it look?
Now this I will not get wrong...

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

$(-1 + 2)^2 + (4-3)^2 = r^2$

$r^2 = 2$

$r = \sqrt{2}$

3. Hello,

Originally Posted by R3ap3r
Find the equation of the circle with center at (-2, 3) and which passes through (-1, 4).

$(x+1)^2 + (y-3)^2=1$

Hows it look?
Hmm...not so good :/

The general equation of a circle is $(x-x_O)^2+(y-y_O)^2=R^2$

Where the center is O ( $x_O;y_O$). Here, the center is (-2,3).

For the radius, replace : you know that (-1,4) is on the circle. So replace x=-1 and y=4 to get R².

4. lol my eye sight fails.

5. $(x+2)^2 + (y-3)^2 = 2$

You think I should write it as 2 or $(\sqrt{2})^2$

6. Originally Posted by R3ap3r
$(x+2)^2 + (y-3)^2 = 2$

You think I should write it as 2 or $(\sqrt{2})^2$
Keep it simple...

7. Originally Posted by R3ap3r
$(x+2)^2 + (y-3)^2 = 2$

You think I should write it as 2 or $(\sqrt{2})^2$
It depends on how you want things to be clear.

The equation of the circle will be the first one.

On your rough copy, it'll be the second one, so that you know that the radius is $\sqrt{2}$

8. gotcha