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A circle has a center at P = (–4, 4) and passes through the point (2,

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Math Expert
Joined: 02 Sep 2009
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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, 05:43
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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.
[Reveal] Spoiler: OA

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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, 08:37
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radius = distance between center and point = $$\sqrt{(2-(-4))^2+(3-4)^2}$$ =$$\sqrt{37}$$

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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Joined: 06 Nov 2014
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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, 03:49
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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, (x-h)^2 + (y-k)^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 (x-h)^2 and (y-k)^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.

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Joined: 02 Sep 2009
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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, 05:14
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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.

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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, 11:44
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, 06:58
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

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At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

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Joined: 23 Jul 2016
Posts: 21
Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]

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10 Nov 2017, 10:14
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.

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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Joined: 23 Jul 2016
Posts: 21
Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]

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15 Nov 2017, 09:02
no one is replying
m getting scared of coordinate geometry now
VP
Joined: 22 May 2016
Posts: 1261
A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]

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15 Nov 2017, 16:40
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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 = x

And here is Bunuel on coordinate geometry

Hope that helps.
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At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

Math Expert
Joined: 02 Sep 2009
Posts: 43387
Re: A circle has a center at P = (–4, 4) and passes through the point (2, [#permalink]

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15 Nov 2017, 19:37
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Psk13 wrote:
:cry: no one is replying
m getting scared of coordinate geometry now

Coordinate geometry and more:

For more check Ultimate GMAT Quantitative Megathread
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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, 22:34
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
Re: A circle has a center at P = (–4, 4) and passes through the point (2,   [#permalink] 01 Dec 2017, 22:34
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