Question

Part 5: Use the information provided to write the standard form equation of each circle.

13. x^2 + y^2 - 26y + 165= 0

14. x^2 + y^2 + 2x + 24y + 140 = 0

15. Ends of a diameter: (-1, -16) and (-1, -10)

Answer 1

13. x² + (y - 13)² = 4

14. (x + 1)² + (y + 12)² = 5

15. (x + 1)² + (y + 13)² = 9

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

Answer 2

The answer to your question is below

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 =
`\sqrt{(-10 + 16)^(2) + (-1 + 1)^(2)}`

-Simplify

dAB =
`\sqrt{6^(2)+ 0^(2)}`

dAB =
`√(36)`

-Result

dAB = 6

radius = 6/2 = 3

-Equation

(x + 1)² + (y + 13)² = 3²

(x + 1)² + (y + 13)² = 9