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What is the least possible distance between a point on the [#permalink]
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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Bunuel wrote: 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 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.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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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 D1 = 12/5 1 = 7/5 = 1.4 we don't require points
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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I have learnt this in my school time that distance from one point to a line is what stated above. Actually its same like finding the line which originates from the distant point and then intersect at the required line, and then we find the point of intersection and then using the formula we calculate distance between the two points. But even if u dont rem the formula what u can do is.... suppose line is ax+byc=0 now when y=0 , x=c/a and when x=0 , y = c/b area of triangle formed by these 2 points and center (0,0) is 1/2 * c/a * c/b = c^2/2ab now this is equal to 1/2 * D1 * D2 , where D1 is distance between points on x and y coordinates of the line which is sqrt [ (c/a)^2 + (b/a) ^2 ] and D2 is the required perpendicular distance on the line. Equate them and u will get the ans. Its very easy concept and I m not that good in explaining here I think sry for that. If you still dont get this please letme know I will explain again.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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14 Dec 2009, 20:55
gurpreetsingh wrote: 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 D1 = 12/5 1 = 7/5 = 1.4
we don't require points I am too weak in this section, could you please explain...what is this formula and why did you do D1? please help



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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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ISBtarget wrote: I am too weak in this section, could you please explain...what is this formula and why did you do D1? please help 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 \(y=\frac{3}{4}*x3\) (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 \((x2,y2)\): \(d=\sqrt{(x1x2)^2+(y1y2)^2}\) 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: \(2.41(radius)=1.4\) 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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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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15 Dec 2009, 13:50
Bunuel wrote: ISBtarget wrote: I am too weak in this section, could you please explain...what is this formula and why did you do D1? please help 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 \(y=\frac{3}{4}*x3\) (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 \((x2,y2)\): \(d=\sqrt{(x1x2)^2+(y1y2)^2}\) 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: \(2.41(radius)=1.4\) 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. Awesome man, why wouldnt you start a quant training program....excellent



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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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17 Dec 2009, 13:55
Hi
One more question...sounds silly but can you help
why are you doing D1 , 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



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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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17 Dec 2009, 14:11
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 xy Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\) 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 \(y = \frac{{3}}{{4}}x3\). 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: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) > \(\frac{height}{3}=\frac{4}{5}\) > \(height=2.4\). \(Distance=heightradius=2.41=1.4\) Answer: A. Hope it's helps.
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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31 Dec 2010, 08:45
Have questions based on this concept (i.e. Circle on a plane) ever been tested by the GMAT?
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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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 (discussed in Veritas Geometry book) Attachment:
Ques2.jpg [ 8.1 KiB  Viewed 20331 times ]
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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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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31 Dec 2010, 10:23
VeritasPrepKarishma wrote: 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 (discussed in Veritas Geometry book) 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. Yes, this approach is quite handy. Question about this concept: triangles106177.html
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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01 Jan 2011, 03:08
Bunuel wrote: Thanks. I'm just reading the GMATclub Coordinate Geometry post now...Great stuff!! It's surprising that the OG Quant book doesn't remotely touch upon the 'Circle on a plane' topic even though it is tested!
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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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Bunuel wrote: ISBtarget wrote: I am too weak in this section, could you please explain...what is this formula and why did you do D1? please help 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 \(y=\frac{3}{4}*x3\) (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 \((x2,y2)\): \(d=\sqrt{(x1x2)^2+(y1y2)^2}\) 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: \(2.41(radius)=1.4\) 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. Brilliant explanation.



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Re: Tough and tricky 7: distance between the circle an line [#permalink]
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06 Dec 2012, 12:45
Bunuel wrote: 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 xy Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\)
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 \(y = \frac{{3}}{{4}}x3\).
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: \(\frac{height}{leg_1}=\frac{leg_2}{hypotenuse}\) > \(\frac{height}{3}=\frac{4}{5}\) > \(height=2.4\).
\(Distance=heightradius=2.41=1.4\)
Answer: A.
Hope it's helps. Hi Bunuel, I could not understand the last but one step. how did you take the ratios?



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Took quite some time(6+ mins) to solve . But I took the same approach as explained by VeritasPrepKarishma. I just hope these ideas click on time. Nice question.
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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 3a) 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. 3b) 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.
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