Difficult Math Question #10

1 !!

Drop a Perpendicular from origin (which is also the center of the circle) on the line. The slope will be -1/m1 where m1 is the slope of the line given.
Get the coordinates
Find the distance X (from the origin)
so frm center it will be ... X- Radius of the circle which is 1

Ans: The equation of the line will be 3x - 4y - 12 = 0.
This crosses the x and y axis at (0,-3) and (4,0)

The circle has the origin at the center and has a radius of 1 unit.

So it is closest to the given line when, a perpendicular is drawn to the line, which passes through the origin.

This distance of the line from the origin is 12 / sqrt (9 + 16) which is 2.4
[Length of perpendicular from origin to line ax +by + c = 0 is

mod (c / sqrt (a^2 + b^2))]


The radius is 1 unit.

So the shortest distance is 2.4 - 1 unit = 1.4 units

can anyone give another explanation for this --a graphical one perhaps because i'm having trouble solving this?

thanks

800guy wrote:i'll post OA when a few people have responded..

What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

also got 1.4
it is easy to draw circle with R=1, and center with the origin
then construct line y=3/4(x)-3
it intersect line x in point (4,0) and line y in point (0,-3)
and forms triangle with sides x,y-lines and y=3/4*x-3

now let us drop perpendicular on the base of triangle from the origin
it will be the height of triangle
the height of triangle-radius of circle is out target
to find the height of triangle different way exist, i solved using area of triangle
Area -1/2*3*4=6
and one more way to find =1/2*base* height
base^2=3^2+4^2=5
area=1/2*5*height
6=1/2*5*h
h=12/5
and r=1
12/5-1=7/5=1/4
hope it makes sence

thanks clock
it makes perfect sense