prove that circle a with center (–1, 1) and radius 1 is similar to circle b with center (–3, 2) and radius 2.

Circle A and circle B are similar

Step-by-step explanation:

* Lets explain similarity of circles

- Figures can be proven similar if one, or more, similarity transformations

 reflections, translations, rotations, dilations can be found that map one

 figure onto another

- To prove all circles are similar, a translation and a scale factor from a

  dilation will be found to map one circle onto another

* Lets solve the problem

∵ Circle A has center (-1 , 1) and radius 1

∵ The standard form of the equation of the circle is:

   (x - h)² + (y - k)² = r² , where (h , k) are the coordinates the center

   and r is the radius

∴ Equation circle A is (x - -1)² + (y - 1)² = (1)²

∴ Equation circle A is (x + 1)² + (y - 1)² = 1

∵ Circle B has center (-3 , 2) and radius 2

∴ Equation circle B is (x - -3)² + (y - 2)² = (2)²

∴ Equation circle B is (x + 3)² + (y - 2)² = 4

- By comparing between the equations of circle A and circle B

# Remember:

- If the function f(x) translated horizontally to the right  

 by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left  

 by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up  

 by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down  

 by k units, then the new function g(x) = f(x) – k

∵ -3 - -1 = -2 and 2 - 1 = 1

∴ The center of circle A moves 2 units to the left and 1 unit up to

  have the same center of circle B

∴ Circle A translate 2 units to the left and 1 unit up

∵ The radius of circle A = 1 and the radius of circle B = 2

∴ Circle A dilated by scale factor 2/1 to be circle B

∴ Circle B is the image of circle A after translation 2 units to the left

  and 1 unit up followed by dilation with scale factor 2

- By using the 2nd fact above

∴ Circle A and circle B are similar

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.