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A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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26 Feb 2015, 06:43
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A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass? A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1) Kudos for a correct solution.
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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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26 Feb 2015, 09:37
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radius = distance between center and point = \(\sqrt{(2(4))^2+(34)^2}\) =\(\sqrt{37}\)
use this distance formula on all the the options they should have same distance. option D. (–3, –2) has the same distance Answer is D.



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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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28 Feb 2015, 04:49
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Bunuel wrote: A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1)
Kudos for a correct solution. Since this is a circle, the distance between any point and the radius will always be the same. Equation of circle is given by formula, (xh)^2 + (yk)^2 = r^2 where (h,k) is center of circle. Hence r^2 = (2+4)^ 2+ (3  4)^2 So, r^2 = 37 So, r = sqrt(37) This should be valid for any other point as well. Since we use terms like (xh)^2 and (yk)^2 i.e. squaring the expression, let us first try the option D because it contains the same numerical values as the point given in the question. So, r^2 = (3+4)^ 2+ (2  4)^2 = 37 Hence this is the point on circle. Hence option D.  Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimusprep.com/gmatondemandcourse



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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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02 Mar 2015, 06:14
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Bunuel wrote: A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1)
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:For this problem, there’s a long tedious way to slog through the problem, and there’s a slick elegant method that gets to the answer in a lightning fast manner. The long slogging approach — first, calculate the distance from (–4, 4) to (2, 3). As it happens, that distance, the radius, equals \(\sqrt{37}\) . Then, we have to calculate the distance from (–4, 4) to each of the five answer choices, and find which one has also has a distance of \(\sqrt{37}\) — all without a calculator. The slick elegant approach is as follows. The point (–4, 4) is on the line y = –x, so it is equidistant from any point and that point’s reflection over the line y = –x. The reflection of (2, 3) over the line y = –x is (–3, –2). Since (–3, –2) is the same distance from (–4, 4) as is (2, 3), it must also be on the circle. Answer = D.
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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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28 Aug 2017, 12:44
The point (–4, 4) is on the line y = –x, so it is equidistant from any point and that point’s reflection over the line y = –x. The reflection of (2, 3) over the line y = –x is (–3, –2). Since (–3, –2) is the same distance from (–4, 4) as is (2, 3), it must also be on the circle. Answer = D.



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A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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30 Aug 2017, 07:58
Bunuel wrote: A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1)
Kudos for a correct solution. I didn't catch shortcut y = x, which IS elegant. I'm not a fan of the distance formula. Start with general formula for a circle: \((x  h)^2 + (y  k)^2 = r^2\). Plug in center coordinates: \((x + 4)^2 + (y  4)^2 = r^2\). Any point through which the circle passes will give you the radius squared. Plug in given point (2,3) \((2 + 4)^2 + (3  4)^2 = r^2\) \(6^2 + (1)^2 = r^2\) \(36 + 1 = 37 = r^2\) Don't solve for \(r\) in this problem with this method. You don't need it. In the answer choices, we need x and y coordinates, that, when plugged into the equation, sum to 37 (satisfy the equation). The circle will pass through that point. Scanning answer choices, because equation contains +4 and 4 and squares, look for values similar to (2,3) and test them [Answer D's (3,2)]. Or do the math for each choice, which is quick when using the circle equation. A. (1,1): \(5^2 + (3)^2 = 34\) NO B. (1,7): \((3)^2 + 4^2 = 25\) NO C. (–1, 9): \((3)^2 + 5^2 = 34\) NO D. (–3, –2): \((1)^2 + (6)^2 = 37\) YES E. (–9, 1): \((5)^2 + (3)^2 = 34\) NO Answer D
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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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10 Nov 2017, 11:14
Bunuel wrote: Bunuel wrote: A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1)
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:For this problem, there’s a long tedious way to slog through the problem, and there’s a slick elegant method that gets to the answer in a lightning fast manner. The long slogging approach — first, calculate the distance from (–4, 4) to (2, 3). As it happens, that distance, the radius, equals \(\sqrt{37}\) . Then, we have to calculate the distance from (–4, 4) to each of the five answer choices, and find which one has also has a distance of \(\sqrt{37}\) — all without a calculator. The slick elegant approach is as follows. The point (–4, 4) is on the line y = –x, so it is equidistant from any point and that point’s reflection over the line y = –x. The reflection of (2, 3) over the line y = –x is (–3, –2). Since (–3, –2) is the same distance from (–4, 4) as is (2, 3), it must also be on the circle. Answer = D. Bunuel can you explain slick elegant approach in detail i am not able to understand why the point reflection is equidistant ? also please explain the line y= x



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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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15 Nov 2017, 10:02
no one is replying m getting scared of coordinate geometry now



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A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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Psk13 wrote: :cry: no one is replying m getting scared of coordinate geometry now Psk13 , if you only ask Bunuel , rest assured that I will never answer. There are legends around here. He is one of them. Now, if you add, "or anyone," or name a bunch of the other terrific experts, also legendary, you're likely to get an answer. (And if you don't, ask again, exactly as you did.) Here is a whole post on the elegance of y = x. This very problem is discussed by a GMATclub expert, mikemcgarry: See here, on y = xAnd here is Bunuel on coordinate geometryHope that helps.
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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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01 Dec 2017, 23:34
Psk13 wrote: :cry: no one is replying m getting scared of coordinate geometry now The coordinates given are (4,4). Here x=4 and y=4 or y=x



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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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11 Feb 2018, 07:35
generis wrote: Bunuel wrote: A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1)
Kudos for a correct solution. I didn't catch shortcut y = x, which IS elegant. I'm not a fan of the distance formula. Start with general formula for a circle: \((x  h)^2 + (y  k)^2 = r^2\). Plug in center coordinates: \((x + 4)^2 + (y  4)^2 = r^2\). Any point through which the circle passes will give you the radius squared. Plug in given point (2,3) \((2 + 4)^2 + (3  4)^2 = r^2\) \(6^2 + (1)^2 = r^2\) \(36 + 1 = 37 = r^2\) Don't solve for \(r\) in this problem with this method. You don't need it. In the answer choices, we need x and y coordinates, that, when plugged into the equation, sum to 37 (satisfy the equation). The circle will pass through that point. Scanning answer choices, because equation contains +4 and 4 and squares, look for values similar to (2,3) and test them [Answer D's (3,2)]. Or do the math for each choice, which is quick when using the circle equation. A. (1,1): \(5^2 + (3)^2 = 34\) NO B. (1,7): \((3)^2 + 4^2 = 25\) NO C. (–1, 9): \((3)^2 + 5^2 = 34\) NO D. (–3, –2): \((1)^2 + (6)^2 = 37\) YES E. (–9, 1): \((5)^2 + (3)^2 = 34\) NO Answer D Hi generis i understand how you got 37, (37 is what ? ) distance ? distance between which points i dont understand where from did you get various values that you plgged in answe6 choices, see below in RED A. (1,1): \(5^2 + (3)^2 = 34\) NO B. (1,7): \((3)^2 + 4^2 = 25\) NO C. (–1, 9): \((3)^2 + 5^2 = 34\) NO D. (–3, –2): \((1)^2 + (6)^2 = 37\) YES E. (–9, 1): \((5)^2 + (3)^2 = 34\) NO Also which formula are you using when testing answer choices ? i know this formula \((x−a)^2\)+\((y−b)^2\) =\(r^2\) By the way did you try to draw this "A circle has a center at P = (–4, 4) and passes through the point (2, 3)." i drew and it doesnt look lik circle with these if center is P = (–4, 4) circle cant pass through (2, 3) , if circle wants to stay a circle and some distorted shape of circle



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A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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dave13 wrote: generis wrote: Bunuel wrote: A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
A. (1, 1) B. (1, 7) C. (–1, 9) D. (–3, –2) E. (–9, 1)
Hi generis i understand how you got 37, (37 is what ? ) distance ? distance between which points i dont understand where from did you get various values that you plgged in answe6 choices, see below in RED A. (1,1): \(5^2 + (3)^2 = 34\) NO B. (1,7): \((3)^2 + 4^2 = 25\) NO C. (–1, 9): \((3)^2 + 5^2 = 34\) NO D. (–3, –2): \((1)^2 + (6)^2 = 37\) YES E. (–9, 1): \((5)^2 + (3)^2 = 34\) NO Also which formula are you using when testing answer choices ? i know this formula \((x−a)^2\)+\((y−b)^2\) =\(r^2\) By the way did you try to draw this "A circle has a center at P = (–4, 4) and passes through the point (2, 3)." i drew and it doesnt look lik circle with these if center is P = (–4, 4) circle cant pass through (2, 3) , if circle wants to stay a circle and some distorted shape of circle @dave13 : (37 is what ? ) distance ? distance between which points 37 is the length of the radius, squared.The radius is the distance between the center of a circle and any point that lies on the circle. (The length of the radius is \(\sqrt{37}\)) where from did you get various values that you plugged in answer choices?I wrote the equation for THIS circle first (from the "standard" formula you know). Then I found the radius using the given point (2,3). (Try plugging that point into the equation for THIS circle in blue, directly below.) The equation for THIS circle is \((x + 4)^2 + (y  4)^2 = r^2\)Then, in the answers, values I "plugged in" were x and ycoordinates for the points in the answers. I just did not show all the steps. I show full steps below for Answers D and A. You say you know this formula: \((x−a)^2\)+\((y−b)^2\) =\(r^2\) Excellent. Standard Equation of a circle,* which is used for circles whose centers are NOT at (0,0) The center is (a,b). The radius is r. (I used the same equation with (h, k).) That is the formula I used to test the answers. But you have to modify the Standard Formula to fit this particular circle.\((x−a)^2\)+\((y−b)^2\) =\(r^2\) You need a center (given), to find the equation for this particular circle. You need one point on the circle (given) to find the length of the radius, squared. Find equation for THIS circle. Plug the center's coordinates (4,4) into standard circle equation: \((x  (4))^2 + (y  4)^2 = r^2\) \((x + 4)^2 + (y  4)^2 = r^2\)  equation for THIS circle Find the radius. Any point that lies on the circle will satisfy this equation (will make the equation true). That point's coordinates will yield the length of the radius, squared. It is given that (2,3) lies on the circle. Plug in point (2,3): \((2 + 4)^2 + (3  4)^2 = r^2\) \((6)^2 + (1)^2 = r^2\) \(36 + 1 = 37 = r^2\) OR \(r = \sqrt{37}\) The way I solved this problem, you can and should keep the radius's length squared. Why? Because any other point that lies on the circle has to satisfy the equation, and  we're squaring everything! The equation gives us \(r^2\). Why make more work? To find whether a point is on the circle, I have to plug (x,y) into the equation we found above: \((x + 4)^2 + (y  4)^2 = r^2\) Any point on the circle, when I plug its coordinates into the equation, will yield \(37 = r^2\)Let's take the other point on the circle, Answer D: (3, 2). Plug in its coordinates. The result MUST be \(37 = r^2\) \((x + 4)^2 + (y  4)^2 = r^2\) \((3 + 4)^2 + (2  4)^2 = r^2\) \((1)^2 + (6)^2 = r^2\) \(1 + 36 = 37 = r^2\) BINGO If you want to take the extra step, \(\sqrt{r^2}\), that is fine. (That step is often used in the distance formula. I'm not a fan of the distance formula.) Do any of the OTHER answers work? (NO.) \((x + 4)^2 + (y  4)^2 = r^2\). Plug in x and y. Result must be 37 • Answer A) point (1,1)? \((1 + 4)^2 + (1  4)^2 = r^2\) \((5)^2 + (3)^2 = r^2\) \(25 + 9 = r^2\) \(25 + 9 = 34 = r^2\) \(r^2\) should equal 37. Here, \(r^2\) equals 34. NO, this point does not lie on the circle. WRONG RADIUS LENGTH Length of radius for this point? \(r = \sqrt{34}\)True length of radius for the circle? \(r = \sqrt{37}\)Try the other answers. Plug the x and ycoordinates into the equation. Do they result in \(r^2 = 37\) or (longer way) \(r = \sqrt{37}\)? Quote: By the way did you try to draw this "A circle has a center at P = (–4, 4) and passes through the point (2, 3)." i drew and it doesnt look lik circle with these if center is P = (–4, 4) circle cant pass through (2, 3) , if circle wants to stay a circle and some distorted shape of circle Yes, I did, for this answer. The circle you describe  which is indeed a circle  is attached. Attachment:
circequation.png [ 49.21 KiB  Viewed 744 times ]
Hope that helps *In the equation, the radius is the hypotenuse of a rightangled triangle. Length of Leg1 of that right triangle: (x  a) Length of Leg 2 of that right triangle: (y  b) \(Leg1^2 + Leg2^2 = Radius^2\) The equation for a circle works because x and y create a right triangle with one vertex at the center of the circle. The equation is a result of the Pythagorean theorem.
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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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14 Feb 2018, 11:07
Many thanks generis for unique explanation one question why did you reduce the font size of the important part of explanation its like in contract where the font size is very small of the most important part of contract



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Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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14 Feb 2018, 12:24
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dave13 wrote: Many thanks generis for unique explanation one question why did you reduce the font size of the important part of explanation its like in contract where the font size is very small of the most important part of contract Agreed about contracts: read the fine print. I put that which you deem the "most important" part of the explanation in small font because I wasn't sure whether it was the most important part. It's hard to know how much people do not know. You said you knew the Standard Formula for a circle. Whether you knew its basis was unclear. I dropped a footnote to address the uncertainty and to keep the post as short as I could. Hint: if you know a formula but do not know why it works or exists, research it.
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A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]
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14 Feb 2018, 13:23
Radius is 6. Hence for y coordinates, the circle will pass through 46=2 & 4+6=10. Since we have an option with 2 as one of the choices, it certainly is D. Though, there can very well be multiple other y coordinates but 2 will always be true as one of the y coordinates.




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