Respuesta :
Answer:
The answer to your question is below
Step-by-step explanation:
The standard form of the equation
(x - h)² + (y - k)² = r²
13.- x² + y² - 26y + 165 = 0
-Group like terms
(x² ) + (y² - 26y ) = -165
-Complete perfect square trinomials
(x² ) + (y² - 26y + 13²) = -165 + 13²
-Simplify
(x² + 0)² + (y - 13)² = -165 + 169
-Result
x² + (y - 13)² = 4
14.
x² + y² + 2x + 24y + 140 = 0
-Group like terms
(x² + 2x ) + (y² + 24y ) = -140
-Complete perfect square trinomials
(x² + 2x + 1²) + (y² + 24y + 12²) = -140 + 1 + 144
-Simplify
(x + 1)² + (y + 12)² = 5
15.-
A ( -1, -16)
B (-1, -10)
-Calculate the midpoint
Xm = (-1 -1) /2 = -1
Ym = (-16 - 10)/2 = -13
Center = (-1, -13)
-Calculate the radius
dAB = [tex]\sqrt{(-10 + 16)^{2} + (-1 + 1)^{2}}[/tex]
-Simplify
dAB = [tex]\sqrt{6^{2}+ 0^{2}}[/tex]
dAB = [tex]\sqrt{36}[/tex]
-Result
dAB = 6
radius = 6/2 = 3
-Equation
(x + 1)² + (y + 13)² = 3²
(x + 1)² + (y + 13)² = 9
Answer:
13. x² + (y - 13)² = 4
14. (x + 1)² + (y + 12)² = 5
15. (x + 1)² + (y + 13)² = 9
Step-by-step explanation:
13. x^2 + y^2 - 26y + 165= 0
h = 0/-2 = 0
k = -26/-2 = 13
0² + 13² - 165 = r²
r² = 4
r = 2
(x - 0)² + (y - 13)² = 2²
x² + (y - 13)² = 4
14. x^2 + y^2 + 2x + 24y + 140 = 0
h = 2/-2 = -1
k = 24/-2 = -12
(-1)² + (-12)² - 140 = r²
r² = 5
(x - (-1))² + (y - (-12))² = 5
(x + 1)² + (y + 12)² = 5
15. Ends of a diameter: (-1, -16) and (-1, -10)
Centre: midpoint of the diameter
(h,k) = (-1-1)/2, (-16-10)/2
= (-1,-13)
Diameter lebgrh= 16 - 10 = 6
Radius = 6/2 = 3
(x - (-1))² + (y - (-13))²= 3²
(x + 1)² + (y + 13)² = 9