chururiakop

2023-03-11

How to identify the center and radius of the circle ${(x+3)}^{2}+{(y-8)}^{2}=16$?

gelo1368m6

Beginner2023-03-12Added 5 answers

Given: ${(x+3)}^{2}+{(y-8)}^{2}=16$

The common shape of a circle $(x-h)}^{2}+{(y-k)}^{2}={r}^{2$, where center is $(h,k)$ and $r$ = radius.

From the given equation ${(x--3)}^{2}+{(y-8)}^{2}=16$ to get a positive value, there must be a negative.

This means the center is $(-3,8)$

Radius $=\sqrt{16}=4$

The common shape of a circle $(x-h)}^{2}+{(y-k)}^{2}={r}^{2$, where center is $(h,k)$ and $r$ = radius.

From the given equation ${(x--3)}^{2}+{(y-8)}^{2}=16$ to get a positive value, there must be a negative.

This means the center is $(-3,8)$

Radius $=\sqrt{16}=4$

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