Question

For the circle x^2 -6x +y^2+8y +5=0 find the center, radius, and the equation of the tangent line at the point (1,-8)

Answer 1

x^2 -6x +y^2+8y +5=0
complete the square:

( x - 3)^2 - 9 + (y + 4)^2 - 16 + 5 = 0

( x - 3)^2 + (y + 4)^2 = 20

compare this with the general form
( x - a)^2 + (y - b)^2 = r^2
where center is (a,b) and radius = r

so from this u can see what center and radius are.

Answer 2

get it?

Answer 3

So the center is 3,-4)?

Answer 4

thats right

Answer 5

And is radical 20 the radius or

Answer 6

radical 20 is right

Answer 7

Should radical 20 stay like that or would 2 radical 5 be a better answer

Answer 8

2 radical 5 is simplest form - yes

Answer 9

and how would you find equation of the tangent?

Answer 10

No idea, I am more confused about that than I am about the center and radius.

Answer 11

So you take your center point and the given point and find the slope and do the inverse of the slope the find slope of tangent? But how do you find the "b" variable of the tangent line

Answer 12

we know coordinates of center C and T is (1, -8)
and the slope of the tangent at T has a slope -1/ slope of CT

Answer 13

use the general form
y-y1 = m(x - x1)

m = slope of tangent and (x1,y1) = (1,-8)

Answer 14

Thanks you so much, I believe I understand how to do it now!

Answer 15

great - yw