SOLUTION: The line {{{x - 2y = -4}}} is tangent to the circle at (0,2). the line {{{y = 2x - 7}}} is tangent to the circle at (3,-1). Find the center of the circle

The line x - 2y = -4 is tangent to the circle at (0,2). the line
y = 2x - 7 is tangent to the circle at (3,-1). Find the center of 
the circle.

We will first draw the picture to see
what's going on, so we'll know what to
do.



The red line is the line x - 2y = -4.
The green line is the line y = 2x - 7

There is a theorem from plane geometry that goes:

If a line is tangent to a circle, then the line is
perpendicular to the radius drawn to the point of 
tangency.

By considering the line of which this radius is a
a segment, we have this corollary:

If a line is tangent to a circle, then the line 
perpendicular to the tangent line passing through 
the point of tangency also passes through the center 
of the circle.

So we find equations of perpendiculars to those two 
tangent lines.  Both willl pass through the center,
and thus their point of intersection must be the 
center of the circle:

First we will find the equation of a line perpendicular
to the red line x - 2y = -4:

First we find its slope by placing it in 
slope-intercept form.  We solve for y:

    x - 2y = -4
       -2y = -x - 4
         y = (1/2)x + 4

Compare that to 
         y = mx + b

and we see that its slope is
m = 1/2.

A line perpendicular to
it will have slope which is the
reciprocal of 1/2 with its sign
changed. Thus we will use m = -2/1
or m = -2.  We want the line
to go through the point of 
tangency (0,2), so we use the
point-slope formula:

y - y1 = m(x - x1)

with m = -2 and (x1,y1) = (0,2)

y - 2 = -2(x - 0)
y - 2 = -2x
    y = -2x + 2

So if we draw in that perpendicular line (in brown), 
we have

The brown line passes through the center of the circle

Now we do the same with the green
line y = 2x - 7. That is, we will 
find the equation of a line
perpendicular to the line y = 2x - 7

y = 2x - 7 is already in  
slope-intercept form. 

Comparing
         y = 2x - 7
to 
         y = mx + b

we see that its slope is
m = 2.

A line perpendicular to
it will have slope which is the
reciprocal of 2 with its sign
changed. Thus we will use 
m = -1/2.  We want the line
to go through the point of 
tangency (3,-1), so we use the
point-slope formula again:

y - y1 = m(x - x1)

with m = -1/2 and (x1,y1) = (3,-1)

y + 1 = (-1/2)(x - 3)
y + 1 = (-1/2)x + 3/2
    y = (-1/2)x + 1/2

Now we will draw that line (in purple), and we have


The brown and purple lines intersect at the center,
so we find their point of intersection by solving
the two equations as a system:

    y = -2x + 2
    y = (-1/2)x + 1/2

To make things easier we multiply the second
equation through by 2

    y = -2x + 2
   2y =  -x + 1

Can you solve this syetem by the substitution method?
If not post again asking how to.

The solution is x = 1, y = 0

So the center of the circle is (1,0)

Edwin