Intersection between unit circle and another circle

the standard form of a circle is 

(x-h)^2 + (y-k)^2 = r^2. 

Center point of the circle is h,k. if (x-2)^2 + (y-5)^2 = 25 then center is (2,5) and the radius of the circle is 5. 

So for the bottom equation you have (x-0)^2 + (y-0)^2 = 1. or at least that is the equivalent equation in standard form. so h,k is 0,0 and the sqrt of 1 is 1. so radius is 1.

for the first equation you have to put it in standard form using complete the square. first group like terms. x^2 + 4x and then y^2 + y, These are binomials that you want to put into trinomial format. In order to complete the square, where a^2 + b^2 + c = 0, then since you currently have a^2+b you take half of b and square it, then add it to both sides.

x^2 + 4x + 4 = 4 and y^2 + y + 1/4 = 1/4. then factor each, which should be (x+2)(x+2) + (y+1/2)(y+1/2) = 3+4+1/4

Now show it in standard form. (x+2)^2 + (y+1/2)^2 = 7+1/4 or 7.25

Thus h = -2, and k = -0.5, with radius as sqroot of 7.25

This is for factoring with missing values as shown in my original answer. 

https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-perfect-squares/v/solving-for-constants-in-perfect-square-polynomial

https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-circle-standard-equation/v/graphing-a-circle-from-standard-equation

and this is graphing circles from standard form as shown in equation #2 you provided. 

Thany you very much. This was very helpful! 

Now I understand how to calculate.

There is just one thing I don't really understand.

I dont understand why the radius is the squareroot of 7.25.

I thought it would be the squareroot of 1.25?

(4 +1/4 -3 = 1.25)?

But you must have calculated:

4+1/4 + 3= 7.25?

I forgot to reverse the sign of the 3 when I move it to the right side of the equation. You sir are correct.