Question 211199
Ok, hello how are you doing today? I am not sure what this question qualifies as under your drop down bar, but I am in college algebra and very stuck on this question. The quesion is about tangent lines and the problem is beneath this statement. 
Write the equation of lines that are tangent to the circle {{{x^2+y^2-6x+2y-16=0}}} when {{{x=2}}}. 
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We find out what points those are by substituting {{{2}}} for {{{x}}} in

{{{x^2+y^2-6x+2y-16=0}}}
{{{2^2+y^2-6(2)+2y-16=0}}}
{{{4+y^2-12+2y-16=0}}}
{{{y^2+2y-24=0}}}
{{{(y+6)(y-4)=0}}}
{{{y+6=0}}} and {{{y-4=0}}}
{{{y=-6}}}  and {{{y=4}}}

So the two points of tangency on the circle are
(2,-6) and (2,4).

We plot those points:

{{{drawing(400,400,-3,9,-6,6, graph(400,400,-3,9,-6,6),

line(2+.1,-6,2-.1,-6), line(2,-6+.1,2,-6-.1), line(2+.1,4,2-.1,4),line(2,4+.1,2,4-.1),locate(2,-6+1,"(2,-6)"), locate(2,4,"(2,4)"), 
line(2+.1,-6+.1,2-.1,-6-.1), line(2+.1,-6-.1,2+.2,-6-.1) )}}} 

We know the circle goes through those two points
and we will draw tangents there:

Next we must find the center of the circle, That is
we must get

{{{x^2+y^2-6x+2y-16=0}}}

in the form

{{{(x-h)^2+(y-k)^2=r^2}}}

We rearrange the terms in the left side,
and get the only constant term on the right

{{{x^2-6x+y^2+2y=16}}} 

Now we want to add two numbers to both sides
in order to complete the squares:

{{{x^2-6x+red("?")+y^2+2y+green("?")=16+red("?")+green("?")}}}

Multiplying the coefficient of x, which is {{{-6}}} by {{{1/2}}}
we get {{{-3}}}, then squaring that gives {{{9}}}.  So we put a 
{{{9}}} where the red question marks are.

Multiplying the coefficient of y, which is {{{2}}} by {{{1/2}}}
we get {{{1}}}, then squaring that gives {{{1}}}.  So we put a 
{{{1}}} where the green question marks are. 

{{{x^2-6x+red(9)+y^2+2y+green(1)=16+red(9)+green(1)}}}

Factor the first three terms on the left, then factor the
last three terms on the left, and combine the terms on the right

{{{(x-3)^2+(y+1)^2=26}}}

So its center is (3,-1) and its radius
is {{{sqrt(26)}}} or about {{{5.099}}}

Now we draw the graph of that circle with that center
and radius and draw in the two radii:

{{{drawing(400,400,-3,9,-6,6, graph(400,400,-3,9,-6,6),
line(2+.1,-6,2-.1,-6), line(2,-6+.1,2,-6-.1), line(2+.1,4,2-.1,4),line(2,4+.1,2,4-.1),locate(2,-6+1,"(2,-6)"), locate(2,4,"(2,4)"),
line(2+.1,-6+.1,2-.1,-6-.1), line(2+.1,-6-.1,2+.2,-6-.1),
circle(3,-1,sqrt(26)), locate(3,-1,"(3,-1)"), line(3,-1,2,-6),line(3,-1,2,4),

line(3+.1,-1,3-.1,-1), line(3,-1+.1,3,-1-.1), line(3+.1,-1,3-.1,-1),line(3,-1+.1,3,-1-.1) 

)}}}

Next we want to find the equations of the two tangent lines
drawn below in light green:

{{{drawing(400,400,-3,9,-7,5, graph(400,400,-3,9,-7,5),
line(2+.1,-6,2-.1,-6), line(2,-6+.1,2,-6-.1), line(2+.1,4,2-.1,4),line(2,4+.1,2,4-.1),locate(2,-6+1,"(2,-6)"), locate(2,4,"(2,4)"),
line(2+.1,-6+.1,2-.1,-6-.1), line(2+.1,-6-.1,2+.2,-6-.1),
circle(3,-1,sqrt(26)), locate(3,-1,"(3,-1)"), line(3,-1,2,-6),line(3,-1,2,4),

line(3+.1,-1,3-.1,-1), line(3,-1+.1,3,-1-.1), line(3+.1,-1,3-.1,-1),line(3,-1+.1,3,-1-.1), 

graph(400,400,-3,9,-7,5,11,(-1/5)x-28/5), graph(400,400,-3,9,-7,5,11,(1/5)x+18/5)

)}}}

The green tangent lines are perpendicular to the 
two radii of the circle drawn above.  You want
the equations of them.  So to finish, you find the 
slopes of the radii, one will be 5 and the other -5,
So you'll use their opposite signed reciprocals,
which will be {{{-1/5}}} and {{{""+1/5}}}. And then
you'll use the point-slope form to find the equations:

They are:

{{{y=(-1/5)x-28/5}}} and{{{y=(1/5)x+18/5}}}

If you can't do that, post again asking how.

Edwin</pre></font></b>