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densk [106]
3 years ago
11

Write the standard equation of a circle that passes through (−2, 10) with center (−2, 6).

Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
6 0
Standard form of a circle" (x-h)²+(y-k)²=r², (h,k) being the center, r being the radius.
in this case, h=-2, k=6, (x+2)²+(y-6)²=r²
use the point (-2,10) to find r: (-2+2)²+(10-6)²=r², r=4
so the equation of the circle is: (x+2)²+(y-6)²=4²
Citrus2011 [14]3 years ago
4 0

Answer:

The equation of the circle is: (x+2)²+(y-6)²= 16

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Answers:

  • Part A) There is one pair of parallel sides
  • Part B) (-3, -5/2) and (-1/2, 5/2)

====================================================

Explanation:

Part A

By definition, a trapezoid has exactly one pair of parallel sides. The other opposite sides aren't parallel. In this case, we'd need to prove that PQ is parallel to RS by seeing if the slopes are the same or not. Parallel lines have equal slopes.

------------------------

Part B

The midsegment has both endpoints as the midpoints of the non-parallel sides.

The midpoint of segment PS is found by adding the corresponding coordinates and dividing by 2.

x coord = (x1+x2)/2 = (-4+(-2))/2 = -6/2 = -3

y coord = (y1+y2)/2 = (-1+(-4))/2 = -5/2

The midpoint of segment PS is (-3, -5/2)

Repeat those steps to find the midpoint of QR

x coord = (x1+x2)/2 = (-2+1)/2 = -1/2

y coord = (x1+x2)/2 = (3+2)/2 = 5/2

The midpoint of QR is (-1/2, 5/2)

Join these midpoints up to form the midsegment. The midsegment is parallel to PQ and RS.

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