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hoa [83]
3 years ago
6

What is the equation of the circle with center (4, 4) that passes through the point (10, 14)? HELP MEEEEE PLEASE

Mathematics
2 answers:
hram777 [196]3 years ago
8 0

Answer:D

Step-by-step explanation:

Gemiola [76]3 years ago
5 0

Answer:

The equation of the circle can be written as:

  • \left(x-4\right)^2+\left(y-4\right)^2=136

Step-by-step explanation:

The general equation of a circle with center (h,k) and radius r is:

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

In our example, we know \left(h,k\right)=\left(4,4\right), as we just have to make sure we need determine \:r^2.

\left(x-4\right)^2+\left(y-4\right)^2=r^2\:\:

As the circle passes through (10, 14), that pair of values for x and y must satisfy the equation. So we have:

\left(10-4\right)^2+\left(14-4\right)^2=r^2

\mathrm{Switch\:sides}

r^2=\left(10-4\right)^2+\left(14-4\right)^2

r^2=6^2+10^2

r^2=36+100

r^2=136

Thus the equation of the circle can be written as:

\left(x-4\right)^2+\left(y-4\right)^2=136

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