least possible distance

[crackgmat007]

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

1.4
sqrt(2)
1.7
sqrt(3)
2

[Katrusya]

I got A

I drew the scheme.
Line and axes form right triangle with sides: 3,4,5.
The least distance between circle and line is height of this right triangle.
Found height = 12/5
then, subtract 1 - radius of the circle
got the answer - 1.4

[crackgmat007]

OA - A

[sanjana]

Can you please explain how height is 12/5?

[aa2kash]

Can you please explain how height is 12/5?

@Sanjana
to find the shortest distance between a point and a line, you need to draw a perpendicular from the point to the line. For example let the point be (h,k)
and an equation of a line is ax+by+c=0.
the distance= |ah+bk+c|/sqrt(a*a + b*b)

in the example above we are trying to find the distance from the centre of the circle i.e from (0,0)
and the equation of theline can be rewritten as
3x-4y-12=0
putting in the above formula you will get
Distance = |3*0 -4*0 -12|/sqrt(3*3 + 4*4)
distance = |-12|/sqrt(25) = 12/5
and once u find the distance from the centre simply subtract the radius to get the distance from a point on the circle which is closest to the line 3x-4y-12=0
Also please not that the point on the circle closest to line is lying on the perpendicular drwan above. thats is why we are subtracting the radius.

Hope its clear to you now.

[CrackGMAC]

Real good application. Thanks again aa2kash

[sanjana]

Perfect aa2kash!
Thanks for ur help

[crackgmat007]

I tried it this way,

we need to find the shortest distance from origin to this line which will be a line from origin perpendicular to this line. In other words, we need to find the hieght of the triangle. Lets take base as 5. Since we know the area, we can use base as 5 & hieght as h. 5 * h * 1/2 = 6. Solving, we get H=2.4. Since the radius of circle is 1, we get 2.4 - 1 = 1.4

[Katrusya]

Great!
I haven't thought to use area formula here, but it is faster and easier.