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What is the least possible distance between a point on the circle x^2

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Re: What is the least possible distance between a point on the circle x^2 [#permalink]

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New post 27 Jul 2015, 13:59
Shree9975 wrote:
Hi Bunuel
Engr2012
Can you please tell why did we took a=-3/4 and b=1
From what we got the values of a and b

Thanks


If you note carefully, that the formula written by Bunuel above for the shortest distance =

Line: \(ay+bx+c=0\), point \((x1,y1)\)

\(d=\frac{|ay1+bx1+c|}{\sqrt{a^2+b^2}}\)

So what it means is that the shortest distance of this line (ax+by+c=0) from (x1,y1) which is the origin (0,0) in our case is

\(d=\frac{|c|}{\sqrt{a^2+b^2}}\). This is obtained by putting x1=0, y1=0,

For a and b, you need to bring the given equation of line :\(y = \frac{3}{4}*x-3\) in the form \(ay+bx+c=0\) to get : \(-4y+3x-12=0\). Comparing these 2 equations we get,

a =-4, b =3 and c = -12

Substitute the above values of a,b,c,x1,y1 in the formula for d above to get:

\(d=\frac{|-12|}{\sqrt{(-3)^2+(4)^2}}=\frac{12}{5}=2.4\) (Bunuel had divided this equation all across by 4. But you can also stick to this method).

And then the final required distance = 2.4 - radius of the circle = 2.4-1 = 1.4.

An alternate solution is below:

Refer to the attached figure for details of the variables used. You can see that the shortest distance to the line will be the perpendicular drawn from the center of the circle.

Also, in triangle POQ, right angled at O and OR is perpendicular to PQ.

In right triangle POQ , \(PQ^2=OP^2+OQ^2\) ---> \(PQ^2 = 3^2+4^2\) ---> \(PQ=5\)

Area of triangle POQ = \(0.5*OP*OQ = 0.5*OR * PQ\) ---> \(OR = \frac{OP*OQ}{PQ}\) ---> \(OR = \frac{3*4}{5}\) ---> \(OR = 2.4\)

Finally,\(OR = OS + SR\), OS = radius of the circle = 1 ---> \(SR = OR - OS = 2.4 -1 = 1.4\). A is the answer.
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Re: What is the least possible distance between a point on the circle x^2 [#permalink]

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New post 30 Apr 2018, 06:25
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Re: What is the least possible distance between a point on the circle x^2   [#permalink] 30 Apr 2018, 06:25

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