Question

the equation of a circle is (x-6)sq +(y+3)sq=36.THe point (8,-3) is on the circle.

What is the equation of the line that is tangent to the circle at (8,-3)?

Answer 1

whats the center of given circle ?

Answer 2

this is all it tells me :(

Answer 3

I got 4 answers to pick from
1- y=8
2- x=8
3- x=-4
y=4

Answer 4

you can tell the `center` of the given circle from its equation

Answer 5

I have to take this final exam in 3 hours

Answer 6

\[\large (x-6)^2 +(y+3)^2=36 \]

\[\large (x-6)^2 +(y-(-3))^2=6^2\]

Answer 7

compare ot wid the standard form of circle :

\[\large (x-h)^2 + (y-k)^2 = r^2\]

center = \((h, k)\)
radius = \(r\)

Answer 8

ok was not sure how to do that part but know I do lol

Answer 9

compare them and tell me what the center is

Answer 10

can you ? :)

Answer 11

would it be 3

Answer 12

center of a circle is a POINT, its not some number okay ?

Answer 13

a POINT can be represented using x and y value : (a, b)

Answer 14

I think its time for bed I am so frustrated :(

Answer 15

\[\large (x-6)^2 +(y-(-3))^2=6^2\]
\[\large (x-h)^2 + (y-k)^2 = r^2\]

Answer 16

just compare above two equations and see if u can tell me below :

\(h = ?\)
\(k = ? \)

Answer 17

this is easy, the only tricky part is finding the center

Answer 18

and finding the center is easy once u know how to find it

Answer 19

by comparing both the equation I get :

\(\large h = 6\)
\(\large k= -3\)

so, center of circle = \(\large (6, -3)\)

Answer 20

I got that part but did not know it was the center

Answer 21

Notice that both `center` and the `given point` have same y value : -3
that means slope = 0

Answer 22

yes

Answer 23

but I get 2 on the other