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djyliett [7]
2 years ago
13

I need help ASAP!!! Please explain how to solve the problem I am stuck

Mathematics
2 answers:
Kazeer [188]2 years ago
7 0

Answer:

( x - 11 )^2 + ( y - 5 )^2 = 4^2

Step-by-step explanation:

The general form of equation for a circle is ( x - h )^2 + ( y - k )^2 = r^2.

h and k are the co-ordinates of the center of circle and r is the radius

We have,

r=4

(h,k)=(11,5)

( x - h )^2 + ( y - k )^2 = r^2

or, ( x - 11 )^2 + ( y - 5 )^2 = 4^2

If you simplify,

x^2-22x+121+y^2-10y+25=16

or, x^2+y^2-22x-10y+146-16=0

or,x^2+y^2-22x-10y+130=0

e-lub [12.9K]2 years ago
3 0

Answer:

( x - 11 )^2 + ( y - 5 )^2 = 4^2

Step-by-step explanation:

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Read 2 more answers
Which of the following points is on a circle if its center is (-13,-12) and a point on the circumference is (-17, -12)?
tia_tia [17]

Answer:

D

Step-by-step explanation:

Obtain the equation of the circle in standard form

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (- 13, - 12), thus

(x + 13)² + (y + 12)² = r²

The radius is the distance from the centre (- 13, - 12) to the point on the circumference (- 17, - 12)

Use the distance formula to calculate r

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 17, - 12) and (x₂, y₂ ) = (- 13, -12)

r = \sqrt{(-13+17)^2+(-12+12)^2} = \sqrt{16} = 4

Hence

(x + 13)² + (y + 12)² = 16 ← in standard form

Substitute the coordinates of each point into the left side of the equation and check

A (- 17, - 13) : (- 4)² + (- 1)² = 16 + 1 = 17 ≠ 16

B (- 9, - 17) : 4² + (- 5)² = 16 + 25 = 41 ≠ 16

C (- 12, 13) : 1² + 25² ≠ 16

D (- 9, - 12) : 4² + 0² = 16

Since (- 9, - 12) satisfies the equation, it is on the circle → D

4 0
3 years ago
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