Question

Discuss the relative fast of the following pairs of circle (x-2)2+(y+1)2=9 , (x-1)2+(y-3)2=4

Answer 1

Answer:

No solutions

Step-by-step explanation:

Let's solve your system by substitution.

(x−2)(2)+(y+1)(2)=9;(x−1)(2)+(y−3)(2)=4

Rewrite equations:

2x+2y−2=9;2x+2y−8=4

Step: Solve2x+2y−2=9for x:

2x+2y−2=9

2x+2y−2+−2y=9+−2y(Add -2y to both sides)

2x−2=−2y+9

2x−2+2=−2y+9+2(Add 2 to both sides)

2x=−2y+11

2x/2=−2y+11/2(Divide both sides by 2)

x=−y+11/2

Step: Substitute−y+11/2

for x in 2x+2y−8=4:

2x+2y−8=4

2(−y+11/2)+2y−8=4

3=4(Simplify both sides of the equation)

3+−3=4+−3(Add -3 to both sides)

0=1

Answer:

No solutions.

Answer 2

Answer:

2x-4+2y+2=8

-2+2x+2y=8

2x+2y=10---1

____&____

2x-2+2y-6=4

2x+2y-8=4

2x+2y=12---2

____&______

2x+2y=10

2x+2y=4

- -. -