This equation has an infinite number of solutions, so it will be categorized as an identity.
Answer:
x^2 + y^2 + 16x + 6y + 9 = 0
Step-by-step explanation:
Using the formula for equation of a circle
(x - a)^2 + (y + b)^2 = r^2
(a, b) - the center
r - radius of the circle
Inserting the values given in the question
(-8,3) and r = 8
a - -8
b - 3
r - 8
[ x -(-8)]^2 + (y+3)^2 = 8^2
(x + 8)^2 + (y + 3)^2 = 8^2
Solving the brackets
( x + 8)(x + 8) + (y +3)(y+3) = 64
x^2 + 16x + 64 + y^2 + 6y + 9 = 64
Rearranging algebrally,.
x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0
Bringing in 64, thereby changing the + sign to -
Therefore, the equation of the circle =
x^2 + y^2 + 16x + 6y + 9 = 0
The equation of line is y = mx+b ; m is slope , b is y- intercept
Please give brainlest
Answer:
(49 - 29) + 1 =(50 - 29)
Step-by-step explanation:
The 20 comes from the subtraction of 29 in both sides of the previous step equality. In the following, I transcript the complete procedure and I add the step that you need to understand why 20 appears (in bold numbers):
29+20=49
49+1=50
(49 - 29) + 1 =(50 - 29)
20+1=21
50-29=21
hence, it was only nesseraty to subtract 29
Answer:
-3(r-2)/2
Step-by-step explanation:
r/
2
−
2
r
+
6
−
3
r
/2
−
2
r⋅
2/
2+
6
−
3
r/
2
+
−
2r
⋅
2/
2
+
6
−
3
r
−
2
r
⋅
2
/2
+
6
⋅
2/
2
+
−
3
⋅
2/
2
r
−
2
r
⋅2
+
6⋅
2
−
3
⋅
2
/2
r
−
4
r
+
12
−
6
/2
−3
(
r
−
2
)
/2
Hope that helps! Plz, mark brainiest! :D
Have a great day!
