Equation of a circle

[Carmen8722]

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

 

[Walter]

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:

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

So you would fill in;

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

 

[Carmen8722]

this answer is

???

Standard equation of a circle is:

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

So you would fill in;

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

 

[Mrspi]

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?

'

 

[Carmen8722]

This answer:

... right??

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?

'

 

[Walter]

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

 

[Mrspi]

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

Walter, I think you (and the original poster!) need to double-check your arithmetic.

[5 sqrt(2)]² is NOT 25.....