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What is the least possible distance between a point on the circle x^2 25 Feb 2021, 00:24
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CEdward
Bunuel
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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. \(2.0\)

Could you please explain the working for arriving at the radius of the circle = 1? Look forward to the replies.

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.

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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You should consider the ratio of corresponding sides. For similar triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).
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What is the least possible distance between a point on the circle x^2 03 May 2021, 09:47
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Glad I could solve this in 10 mins.

I saw that if x was 1 or -1, then y would be 0, and from that I knew the radius was 1. Then I had to calculate the height (h) of the right triangle(s) = the distance from origo to the line.

(5-x)^2 + h^2 = 3^2
x^2 + h^2 = 4^2

-->

3^2 - (5-x)^2 = 4^2 - x^2 -->

x = 3,2

3,2^2 + h^2 = 16^2 -->

h^2 = 16 - 16^2/5^2 -->

h^2 = 16(1 - 16/25) = 16*(9/25) -->

h = 12/5

12/5 - 1 = 7/5 = 1,4

Now I will move on to learn from the magic formula presented here...
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What is the least possible distance between a point on the circle x^2 16 Aug 2021, 08:50
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Bunuel
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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Whats the difference between option A and option B? root 2 = 1.4
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What is the least possible distance between a point on the circle x^2 30 Mar 2022, 08:50
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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.

Whats the difference between option A and option B? root 2 = 1.4
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1.4 means 1.4000 but root 2 means 1.4142 which is more than 1.4000
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What is the least possible distance between a point on the circle x^2 01 Apr 2022, 04:54
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I have few doubts here,

1. A triangle has multiple heights depending on respective base we take, am I correct ?
2. Over here why is it that when we take height as leg 1 (3units) as it is perpendicular to leg2 (4 units) and then subtract radius from it (3-1 = 2 units) why don't we get minimum distance, what ensures that drawing a perpendicular from the origin to the line will get us minimum distance
3. d=|ay1+bx1+c|/sqrt a2+b2, over here it will be correct to take a = 1,b=-3/4, c= 3 or a= -1, b= 3/4, c = -3, both will do right ?
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What is the least possible distance between a point on the circle x^2 01 Apr 2022, 06:06
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KarishmaB
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What is the least possible distance between a point on the circle \(x^2 + y^2 = 1\) and a point on the line \(y = \frac{3}{4}x - 3\) ?

\(1.4\)
\(\sqrt{2}\)
\(1.7\)
\(\sqrt{3}\)
\(2.0\)

Can anyone Please explain this Question??

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)

Attachment:
File.jpg

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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Thankyou for the easy explanation mam :)
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What is the least possible distance between a point on the circle x^2 01 Apr 2022, 07:51
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KarishmaB
CracktheGmat2010
What is the least possible distance between a point on the circle \(x^2 + y^2 = 1\) and a point on the line \(y = \frac{3}{4}x - 3\) ?

\(1.4\)
\(\sqrt{2}\)
\(1.7\)
\(\sqrt{3}\)
\(2.0\)

Can anyone Please explain this Question??

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)

Attachment:
File.jpg

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


Thankyou for the easy explanation mam :)
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shivaniilall

Please refer sr. no. 11 on this link https://gmatclub.com/forum/rules-for-po ... l#p1092822
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What is the least possible distance between a point on the circle x^2 02 Apr 2022, 06:29
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Kushchokhani The solution above really helped me and if I'd see a similar question in exam, I am sure to save two minutes on it. And so, I did not think my comment was unnecessary. I think we can let that be known on this forum.
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What is the least possible distance between a point on the circle x^2 02 Apr 2022, 07:19
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shivaniilall
Kushchokhani The solution above really helped me and if I'd see a similar question in exam, I am sure to save two minutes on it. And so, I did not think my comment was unnecessary. I think we can let that be known on this forum.
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shivaniilall

I am sure you'll come across many such useful solutions in different topics. For this reason, kudos have been introduced. Kudos are a way to give thank you without need to separately post. We understand your sentiments, but imagine if everyone gives a thank you post separately for most of the topics, then the important posts may get lost in the ocean, topics ultimately running in pages after pages. We truly appreciate your thanks in the form of kudos as any person can see who all have given kudos (thanks) for any particular post. Also, kudos have value on GMAT Club in the form of rewards, but thank you post doesn't. No one other than the person to whom we thank is interested to read thank you post and person like me who has a habit to look into detailed discussions will have to go through such kind of posts, thereby not adding any value to our time. So, in the benefit of all, kudos works well.

Your previous post was removed by the moderator for this only reason.

Trust you understand.
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What is the least possible distance between a point on the circle x^2 26 Jul 2022, 17:24
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Bunuel
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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Hi there,
Could you explain how you setup this ratio?

heightleg1=leg2hypotenuseheightleg1=leg2hypotenuse

I'm having trouble understanding it conceptually.
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What is the least possible distance between a point on the circle x^2 01 Jun 2023, 14:35
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KarishmaB
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
Attachment:
Ques2.jpg

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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where did you get 5 from?
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What is the least possible distance between a point on the circle x^2 27 Aug 2025, 08:22
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