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

A rather nice question and I think Bunuel has already explained sufficient alternative solutions. I just want to add a method of finding the altitude of a right triangle here that I think is particularly neat


I want to find x here since (x - 1) will be the minimum distance from the circle to the line (as explained above)

Area of the given right triangle = (1/2)*3*4 (3 is base and 4 is altitude)= (1/2)*5*x (5 is base and x is altitude)
So x = 2.4

Finding the area of the original triangle in two different ways and equating it will help you find the altitude.
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Lets do in 1 min.

Use the formula

D = | Am+Bn+C|/ SQRT(A^2 + B^2) where Ax+By+C = 0

put (m,n) =0,0 = center of circle

we get D = 12/5 thus required distance is D-1 = 12/5 -1 = 7/5 = 1.4

we don't require points
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First note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

Now we can do this by finding the equation of a line perpendicular to given line \(y=\frac{3}{4}*x-3\) (we know it should cross origin and cross given line, so we can write the formula of it), then find the croos point of these lines and then the distance between the origin and this point. But it's very lengthy way.

There is another, shorter one. Though I've never seen any GMAT question requiring the formula used in it.

We know the formula to calculate the distance between two points \((x1,y1)\) and \((x2,y2)\): \(d=\sqrt{(x1-x2)^2+(y1-y2)^2}\) BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:
Line: \(ay+bx+c=0\), point \((x1,y1)\)

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

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is \((0,0)\) -->

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

So in our case it would be: \(d=\frac{|-3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4\)

So the shortest distance would be: \(2.4-1(radius)=1.4\)

Answer: A.

P.S. Also note that when we have \(x^2+y^2=k\), we have circle (as we have \(x^2\) and \(y^2\)), it's centered at the origin (as coefficients of \(x\) and \(y\) are \(1\)) and the radius of that circle \(r=\sqrt{k}\).

You can check the link of Coordinate Geometry below for more.
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In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)

If the circle is centered at the origin (0, 0), then the equation simplifies to:
\(x^2+y^2=r^2\)

So, the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line \(y = \frac{{3}}{{4}}x-3\).

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> \(leg_1=4\).
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> \(leg_2=3\).

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) --> \(\frac{height}{3}=\frac{4}{5}\) --> \(height=2.4\).

\(Distance=height-radius=2.4-1=1.4\)

Answer: A.
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Good problem again. Takes into account a lot of co ordinate geometry fundas.

eqn of circle = x^2 + y^2 = 1, center = (0,0) radius = 1

min dist of line from circle = dist of line from the center - radius

Make the distance of the line from the circle to be 0 and we see that it becomes a tangent to the circle.

Now we know if we draw a line from the center to the point where the tangent touches the circle the line and the tangent are perpendicular to each other.

So we need to find the equation of this line first.

We can take the line back where it was now

Since the lines are perpendicular m1 x m2 = -1

m of line = 3/4

so slope of the new line = -4/3

Since the line passes through the origin (center of circle) its eqn => y=-4/3x

now we need to get the point of intersection of our two lines, which comes out to be (36/25,-48/25)

now get the distance of this point from the origin and subtract the radius from it.

Comes to 1.4 (may have made calculation errors )

So A.

Comes under 2 mins.

Bunuel, great work with the Questions. I suggest you make a single thread and keep updating it. People can subscribe to that and also it will help new guys read all the probs and solutions in one thread.

I am too weak in this section, could you please explain...what is this formula and why did you do D-1?
please help

Awesome man, why wouldnt you start a quant training program....excellent

Hi

One more question...sounds silly but can you help

why are you doing D-1 , you are calculating the distance between a point on the circle and the line , 0,0 is a point on the circle, why cant 2.4 be the answer

Minimum distance from the circle to the line would be:
Length of perpendicular from the origin to the line (as the circle is centered at the origin) - The radius of a circle (which is 1).

(0,0) is not the point on the circle, it's the center of the circle with radius 1 (circle is centered at the origin).

2.4 is the distance from the line to the origin (the center of the circle), so we should subtract the length of the radius to get the distance from the line to the circle.

ANOTHER SOLUTION:

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)

If the circle is centered at the origin (0, 0), then the equation simplifies to:
\(x^2+y^2=r^2\)

So, the circle represented by the equation \(x^2+y^2 = 1\) is centered at the origin and has the radius of \(r=\sqrt{1}=1\).

Then note that min distance from the circle to the line would be: length of perpendicular from the origin to the line (as the circle is centered at the origin) - the radius of a circle (which is 1).

So we should find the length of perpendicular, or the height of the right triangle formed by the X and Y axis and the line \(y = \frac{{3}}{{4}}x-3\).

The legs would be the value of x for y=0 (x intercept) --> y=0, x=4 --> \(leg_1=4\).
and the value of y for x=0 (y intercept) --> x=0, y=-3 --> \(leg_2=3\).

So we have the right triangle with legs 4 and 3 and hypotenuse 5. What is the height of this triangle (perpendicular from right angle to the hypotenuse)? As perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) --> \(\frac{height}{3}=\frac{4}{5}\) --> \(height=2.4\).

\(Distance=height-radius=2.4-1=1.4\)

Answer: A.

Hope it's helps.
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I have a different approach to this. I first checked the options and there is 15-20% differences in the values of all optoins.

From x2 + y2 = 1, we have center at (0,0). Now, for the shortest distance b/w two points is the perpendicular distance.

The slope of line | to line y=3x/4 -3 will be (-4/3) and as this | is in line with the center of circle, so its line eq => y = -4x/3

Now, just equating the two line equations to solve for x and y, we get:
x=36/25 and y=48/25. I assumed the closest values from these ratios as:
x=3/2 and y=2

This is for quick calculation. So, the distance b/w (0,0) and (3/2,2) is nearly \sqrt{(25/4)} is 2.5. We assumed the ratio above (x=36/25 = 3/2 and y=48/25 = 2) a bit more. So, this number should be a bit less. So, I took it as 2.4.

You have noticed that the radius is 1. Thus, the distance b/w a point at circle and the first line will be 2.5 -1 = 1.4, which is A.

I hope this helps.

Thanks ykaiim and Bunnel for the explanations..

Bunnel, you provide some of the best explanations for the Quant problems.. Thanks again!

Equation for line can be written as : 4y-3x+12 = 0
Some basic equations :

Distance between any point ( x1, y1 ) and line ax+by+c=0 is

ax1+by1+c/ \sqrt{a2+b2}

Here (x1,y1 ) is (0,0) ( center of circle )
ANd from equation of line , a=3 b=4 and c =12
Hence distance between center of circle and line will be :

= 4*0 + 3*0 + 12/ \sqrt{4*4 + 3* 3}
= 12/\sqrt{5}
=2.4

Distance between center of circle and line is 2.4 .

SInce radius is 1 , least distance between line and circle will be 2.4-1 = 1.4

Hope this helps

Yes, this question can be solved this way too, though I've never seen any GMAT question requiring the formula used in it. Also there is a little mistake in formula you posted: nominator should be in absolute value.

What is the least possible distance between a point on the circle \(x^2+y^2 = 1\) and a point on the line \(y = \frac{{3}}{{4}}x-3\)?
A. \(1.4\)
B. \(\sqrt{2}\)
C. \(1.7\)
D. \(\sqrt{3}\)
E. \(2.0\)

\(y = \frac{{3}}{{4}}x-3\) --> \(y-\frac{{3}}{{4}}x+3=0\).

We know the formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\): \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) BUT there is a formula to calculate the distance between the point (in our case origin) and the line:

DISTANCE BETWEEN THE LINE AND POINT:
Line: \(ay+bx+c=0\), point \((x_1,y_1)\)

\(d=\frac{|ay_1+bx_1+c|}{\sqrt{a^2+b^2}}\)

DISTANCE BETWEEN THE LINE AND ORIGIN:
As origin is \((0,0)\) -->

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

So in our case it would be: \(d=\frac{|3|}{\sqrt{1^2+(\frac{3}{4})^2}}=\frac{12}{5}=2.4\)

So the shortest distance would be: \(2.4-1(radius)=1.4\)

Answer: A.
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Brilliant explanation.

The question has been discussed before. This is my take on it.

Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? (shown by the bold line)



Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer.

Then, let's try to find the length of the bold line, x.
Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line
x = 2.4
So minimum distance is 2.4 - 1 = 1.4
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For the people still troubled with this questions -

First of all this questions tests relatively advanced skills in mathematics and hence IMO can not be a part of the real GMAT.

However, to find the answer of this Q we need to follow below mentioned steps:

1) find the distance of the line from origin : this distance should be the shortest possible distance
2) as all points on circle are equi distance from the origin, we need to find the shortest distance of line from origin and subtract radius from it to get our answer
3) to get shortest distance we need to actually find length of perpendicular line which starts from origin and ends at our given line

3-a) One of the method to solve for (3) is using the equation - |ax1 + by1 + c| /sqrt (a^2+b^2) (this formula you need to remember) - read posts by bunuel or gurpreet for more details.
3-b) Another method is to find the equation of perpendicular line and then find an intersection point of this perpendicular line with our given line. Now find the distance between this point to origin (PHEW) - I surely can't do all this in less 2 minutes and be accurate to the second decimal point [ remember our options are 1.4 and sqrt (2) = 1.41 ]


Finally, for the purpose of GMAT only, i would advise you should not be worried if you can't remember this formula or find this question too difficult.

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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Bunuel Thanks for this post. It's really important to be clear. I followed what you did until this:

height/leg1=leg2/hypotenuse --> height/3 = 4/ 5

Can you elaborate on this further? What's the general rule for setting up the ratios? This is something that I struggle with.
How does it differ with respect to similar triangles or I am guessing this is the same thinking...
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