Question

If the equation of a circle is (x + 4)2 + (y - 6)2 = 25, its center point is

(4, 6)
(-4, 6)
(4, -6)

Answer 1

Recall that the standard form of a circle is
\[(x-h)^2+(y-k)^2=r^2\] where \((h,k)\) is the center of the circle.

Do you think you can identify the center of the circle in your problem now? :-)

Answer 2

not really beacuse i get confused how to do it

Answer 3

Rewrite your equation as \(\large (x-(-4))^2 + (y-6)^2 = 25\). Now when you compare this equation to the standard form I provided above, what is (h,k)? I'll put the two equations side by side to make it easier to see what I'm getting at.
\[\large \begin{aligned}(x-h)^2+(y-k)^2 &= r^2\\ (x-(-4))^2+(y-6)^2 &= 25\end{aligned}\]

Answer 4

okay sence h is 4 and k is 6 how do i get my answer

Answer 5

is the answer (4, 6)

Answer 6

careful, when you compare \(\large (x-h)^2\) with \(\large (x-(-4))^2\), you should see that \(\large h=-4\). You had the right value for k, so it turns out that the center of the circle is (-4,6),

Does this clarify things? :-)

Answer 7

yes it helps but i dont 100% get it but thanks you for helpng me out