A circle of radius 4 touches the line 3x+4y=8. Find the equation of the center of the circle.?

Question

plz..help

pls..show how ur able to arrive at that answer..thanks..

Answer 1

Just sketch the diagram and you will see
there is almost no calculation
involved.

This is an equation y= -3/4 x +2
y-intersect is 2.
There are infinite # of circles touching the line.
Assume a circle is touching this point.
Draw the radius from (0,2) to the center(x,y)

As you are drawing the radius with a
perpendicular line to the center
the slope of this radius is 4/3.
You will find it forms a 3-4-5 triangle.
When you draw the parallel line of center of the circles,
you will see this line intersects
the y-axis at (0,7).
The parallel line between (0,7) and
the center (x,y) is the line of the center of the circle
you are looking for.
The slope is -3/4, y-intercept is 7.
So the equation is
y = -3/4 x +7
The line of circle is shifted 5 units up.

But the circle can also be on the other
side of the line.
So another parallel line will be 5 units down
from the original y-intercept (0,2),
to (0,-3)
So the second equation is
y=-3/4x -3

Answer 2

3+4y'=0
y'=-3/4

3a+4b=8
radius slope is 4/3

A distance from center(a,b) to the tangent line 3x+4y-8=0
is
(3a+4b-8)/sqrt(9+16)=4
Hence we have two equations with two unknowns a and b.
Solve them find a, and b

Answer 3

If (a,b) are the center coordinates
(3a+4b-8)/sqrt(9+16) = +-4
so
3a+4b=8+-20