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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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A perpendicular line has an opposite and a reciprocal of the slope.

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Since the perpendicular line has an opposite and a reciprocal of the slope, the slope will be -5/3.

Now you must make an equation in point-slope form. This is an example of that form. You will need at least one point to make this equation work. In this case we have (-8,0).

In put the y and x coordinates like this:

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This is your equation.

y = -5/3 * x - 40/3

(You can make it -5/3x in your answer but it looks weird online. You may think that it is -5 divided by 3 times x, but it actually is 5/3 times x. That's why I wrote it as y = -5/3 * x - 40/3)

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