Question

The center of a circle is at (-10, 6) and it has a radius of 4. What is the equation of the circle?
(Points : 4)
(x -10)^2 + (y + 6)^2 = 2

(x -10)^2 + (y + 6)^2 = 16

(x + 10)^2 + (y - 6)^2 = 2

(x + 10)^2 + (y - 6)^2 = 16

Answer 1

are you aware of how to find the equation or a circle?

Answer 2

using the radius right?

Answer 3

\[\huge\rm (x-h)^2 + (y-k)^2 = r^2\]
standard form equation of circle
where (h,k) is the center and r is radius

Answer 4

Oh yeah that does look really familiar

Answer 5

okay so replace (h,k) points by (-10,6)
r=4

Answer 6

An equation of the circle with center \(\large\color{black}{ \left(\color{red}{\rm h},\color{blue}{\rm k}\right) }\) and radius \(\large\color{green}{ \rm r }\) is:


\(\large\color{black}{ (x-\color{red}{\rm h})^2+(y-\color{blue}{\rm k})^2=\color{green}{\rm r}^2 }\)

Answer 7

took a little long to do the colors

Answer 8

so its( x- 10^2 )+ (y- 6)^2 = 4

Answer 9

^2 is outside the parenthesis

Answer 10

r=4
r^2 = ?

Answer 11

oops it would be after that whats next?

Answer 12

^2 next to the 4 on the right side.

Answer 13

@Nnesha 16

Answer 14

yes right

Answer 15

Then?

Answer 16

re-write it correctly please. with all corrections

Answer 17

(x-10)^2 + (y-6)^2 = 16

Answer 18

yes

Answer 19

good job, bye

Answer 20

wait i still need help figuring it out

Answer 21

\(\color{blue}{\text{Originally Posted by}}\) @cookiimonster27
(x-10)^2 + (y-6)^2 = 16
\(\color{blue}{\text{End of Quote}}\)
nope
there is a negative sign in the parentheses and h is also negative so...

Answer 22

itd be (x-(-10))^2 + (y-6)^2=16 @Nnesha which would make the 10 positive so its D

Answer 23

yes right

Answer 24

can you help with one last ine please

Answer 25

hmm i'll try

Answer 26

The equation of a parabola is shown.
y= 1/10x^2
What are the coordinates of the focus?
(Points : 4)
(0, 2.5)

(0, -2.5)

(0, 0.8)

(0, -0.4)
I think its D

Answer 27

it's horizontal or vertical ?

Answer 28

It goes like that

Answer 29

okay so it's ? vertical or horizontal ? :-)

Answer 30

remember the quadratic equation
y =Ax^2+Bx+C
^^^ x is squared so

Answer 31

I think vertical

Answer 32

yes right x is squared so it's vertical

Answer 33

Yes x is squared .what do you think it is

Answer 34

so focus should be \[\huge\rm (0 , \pm P)\]
do you know wat is p?

Answer 35

I thhnk its 0.04

Answer 36

okay standard form is
4p should be with y
so that's why first solve for x^2