# Thread: Equation of a circle

1. ## Equation of a circle

How to find and solve the equation of a circle with radius 5 radical 2 and center (-3,2)???

2. Originally Posted by Carmen8722
How to find and solve the equation of a circle with radius 5 radical 2 and center (-3,2)???
Standard equation of a circle is:

$(X-H)^2 + (Y-K)^2 \:=\: R^2$

So you would fill in;

$(X + 3)^2 + (Y - 2)^2 \:=\: 25$

Originally Posted by Walter
Standard equation of a circle is:

$(X-H)^2 + (Y-K)^2 \:=\: R^2$

So you would fill in;

$(-3 - H)^2 + (2 - K)^2 \:=\: 25$

4. Originally Posted by Carmen8722
How to find and solve the equation of a circle with radius 5 radical 2 and center (-3,2)???
The general form for the equation of a circle with center (h, k) and radius r is

(x - h)2 + (y - k)2 = r2

You are GIVEN the center: (-3, 2) so if (h, k) = (-3, 2) then h = -3 and k = 2.

You are GIVEN that the radius is 5 sqrt(2)...that's r.

The equation of the circle in standard form, then, is

(x - (-3))2 + (y - 2)2 = [5 sqrt(2)]2

Can you do some simplification?

'

Originally Posted by Mrspi
The general form for the equation of a circle with center (h, k) and radius r is

(x - h)2 + (y - k)2 = r2

You are GIVEN the center: (-3, 2) so if (h, k) = (-3, 2) then h = -3 and k = 2.

You are GIVEN that the radius is 5 sqrt(2)...that's r.

The equation of the circle in standard form, then, is

(x - (-3))2 + (y - 2)2 = [5 sqrt(2)]2

Can you do some simplification?

'

6. He's right, I got the equation mixed up, the answer would be;
$(X + 3)^2 + (Y - 2)^2 \:=\: 25$

7. Originally Posted by Walter
He's right, I got the equation mixed up, the answer would be;
$(X + 3)^2 + (Y - 2)^2 \:=\: 25$
Walter, I think you (and the original poster!) need to double-check your arithmetic.

[5 sqrt(2)]2 is NOT 25.....

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