Question

1. Find the equation of the tangent lines to the circle x^2+ y^2 + 6x - 2y+5=0, perpendicular to the line 2x +y = 4. Draw the Figure.

Answer 1

@dan815

Answer 2

do you get to use calculus or no? if no, it is really a pain and i forget how to do it

Answer 3

its my analytic geometry dude.. its my major subject, and yet i cant solve it D:

Answer 4

yeah i remember these, quite annoying
something about the line touching at one point so the quadratic equation has one solution so the discriminant is zero blah blah

Answer 5

you are looking for a line with slope \(\frac{1}{2}\) that is tangent to the curve
with calc it is a no brainer i think, without it probably a google search

Answer 6

i think OS is broken, i cant post my messages

Answer 7

it is just slow is all

Answer 8

they improve it to be slow

Answer 9

so do you know how to work with this problem?

Answer 10

maybe we can figure this out
do you have a worked out example or not?

Answer 11

nope, nothing similar with it

Answer 12

if there is, i can work it by myself

Answer 13

dang i totally have seen these but it is so much easier with calc that i can't remember
the circle has center \((-3,1)\) and radius \(\sqrt5\) which you get by completing the square and writing it as
\[(x+3)^2+(y-1)^2=5\]

Answer 14

yup im stuck at that part

Answer 15

so you have to find a point on the circle with tangent line \(\frac{1}{2}\)
it will be perpendicular to the line through the center with slope \(-2\) so maybe we can find the intersection of the line with slope \(-2\) though \((-3,1)\) and see where it intersects the circle
that is the point you are looking for

Answer 16

the line is
\[y-1=-2(x-3)\] or
\[y=-2x+7\]

Answer 17

plug that in to the equation for the circle i.e. replace \(y\) by \(-2x+7\) and solve the resulting equation
that should work

Answer 18

nope scratch that, it has no solutions which is a little confusing

Answer 19

you just substitute that, is it possible?

Answer 20

hold on i think i screwed up the line

Answer 21

\[y-1=-2(x+3)\\
y-1=-2x-6\\
y=-2x-5\] jeez my algebra stinks

Answer 22

that will work now

Answer 23

\[(x+3)^2+(y-1)^2=5\\
(x+3)^2+(-2x-5-1)^2=5\\
(x+3)^2+(-2x-6)^2=5\]

Answer 24

solve that quadratic equation for \(x\)
that will give you two x values for the point of intersection

Answer 25

in fact, they are nice integers so i am sure the answer is right
check here for the two points of intersection
http://www.wolframalpha.com/input/?i=+x^2%2B+y^2+%2B+6x+-+2y%2B5%3D0%2C+y%3D-2x-5

Answer 26

when you expand that nonsense out, you get
\[5 x^2+30 x+40 = 0\]
divide by 5, then factor (if you have to show all your work)

Answer 27

you good from there?

Answer 28

x^2 +6x+8=0

Answer 29

yeah which factors nicely

Answer 30

so i will use QF to do it

Answer 31

nah just
\[(x+2)(x+4)=0\]

Answer 32

i think we need to get the 2 lines

Answer 33

yeah you get two different values for \(x\) namely \(-2\) and \(-4\)