I will explain you and pair two of the equations as an example to you. Then, you must pair the others.
1) Two circles are concentric if they have the same center and different radii.
2) The equation of a circle with center xc, yc, and radius r is:
(x - xc)^2 + (y - yc)^2 = r^2.
So, if you have that equation you can inmediately tell the coordinates of the center and the radius of the circle.
3) You can transform the equations given in your picture to the form (x -xc)^2 + (y -yc)^2 = r2 by completing squares.
Example:
Equation: 3x^2 + 3y^2 + 12x - 6y - 21 = 0
rearrange: 3x^2 + 12x + 3y^2 - 6y = 21
extract common factor 3: 3 (x^2 + 4x) + 3(y^2 -2y) = 3*7
=> (x^2 + 4x) + (y^2 - 2y) = 7
complete squares: (x + 2)^2 - 4 + (y - 1)^2 - 1 = 7
=> (x + 2)^2 + (y - 1)^2 = 12 => center = (-2,1), r = √12.
equation: 4x^2 + 4y^2 + 16x - 8y - 308 = 0
rearrange: 4x^2 + 16x + 4y^2 - 8y = 308
common factor 4: 4 (x^2 + 4x) + 4(y^2 -8y) = 4*77
=> (x^2 + 4x) + (y^2 - 2y) = 77
complete squares: (x + 2)^2 - 4 + (y - 1)^2 - 1 = 77
=> (x + 2)^2 + (y - 1)^2 = 82 => center = (-2,1), r = √82
Therefore, you conclude that these two circumferences have the same center and differet r, so they are concentric.
Answer:
What is the arc length and sector area for the following circle. Round your answer to 4 decimal places.
Step-by-step explanation:
Answer:
Q1. x= 18, y=59
Q2. m∠J= 56°
Step-by-step explanation:
Q1. (3x +5)°= y° (base ∠s of isos. △)
y= 3x +5 -----(1)
(3x +5)° +y° +(4x -10)°= 180° (∠ sum of △)
3x +5 +y +4x -10= 180
7x +y -5= 180
7x +y= 180 +5
7x +y= 185 -----(2)
Substitute (1) into (2):
7x +3x +5= 185
10x= 185 -5
10x= 180
x= 180 ÷10
x= 18
Substitute x= 18 into (1):
y= 3(18) +5
y= 59
Q2. (5x -13)°= (3x +17)° (base ∠s of isos. △)
5x -13= 3x +17
5x -3x= 17 +13
2x= 30
x= 30 ÷2
x= 15
∠LKJ
= 3(15) +17
= 62°
∠KLJ= 62° (base ∠s of isos. △)
m∠J
= 180° -62° -62° (∠ sum of △JKL)
= 56°