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Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin

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Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 11 Sep 2018, 06:45
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Question Stats:

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[GMATH practice question]

Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4

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Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 11 Sep 2018, 07:01
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fskilnik wrote:
[GMATH practice question]

Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4



when any line or curve intersects vertical or Y-axis, the value of x is 0..
so substitute x as 0
\((x-3)^2 + (y-1)^2 = 13.......(0-3)^2 + (y-1)^2 = 13......(y-1)^2=13-9=4\)
so two cases
1. \(y-1=2....y=3\)
2. \(y-1=-2...y=-1\)

distance = \(p = 3-(-1)=3+1=4\)

E
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Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 11 Sep 2018, 07:04
fskilnik wrote:
[GMATH practice question]

Circle \((x-3)^2 + (y-1)^2 = 13\) ntersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4


On the vertical axis, x=0.
So, the intersection points of the circle with vertical axis are (0,\(y_{1}\)) and (0,\(y_{2}\))

Now, put x=0 in the equation of the given circle to get the values of \(y_{1}\) and ,\(y_{2\)

Hence, \((0-3)^2+(y-1)^2=13\)
or, \(9+(y-1)^2=13\)
or, \((y-1)^2=4\)
Or, y-1=2, y-1=-2
Or, \(y_{1}\)=3, \(y_{2}\)=-1

p=3-(-1)=4

Ans.(E)
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Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 11 Sep 2018, 08:06
OA: E

The equation of circle : \((x-3)^2 + (y-1)^2 = 13\)

Centre of circle :\((3,1)\)

\({Radius}^2 \quad=13\)
Attachment:
Circle.PNG
Circle.PNG [ 14.81 KiB | Viewed 629 times ]


Using Pythagoras theorem, we get

Required Chord length \(= 2* \sqrt{{{Radius}^2-3^2}}= 2* \sqrt{{13-9}}=2* \sqrt{{4}}=4\)
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Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 11 Sep 2018, 19:52
chetan2u wrote:
fskilnik wrote:
[GMATH practice question]

Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4



when any line or curve intersects vertical or Y-axis, the value of x is 0..
so substitute x as 0
\((x-3)^2 + (y-1)^2 = 13.......(0-3)^2 + (y-1)^2 = 13......(y-1)^2=13-9=4\)
so two cases
1. \(y-1=2....y=3\)
2. \(y-1=-2...y=-1\)

distance = \(p = 3-(-1)=3+1=4\)

E


In cases like these, to find distance, how will we choose coordinates as Y1 and Y2? Like in this problem, while taking the difference b/w y-coordinates, how have you chosen the y coordinates? Cant it be -1-3=-4?
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Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 11 Sep 2018, 19:59
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Harikamini wrote:


In cases like these, to find distance, how will we choose coordinates as Y1 and Y2? Like in this problem, while taking the difference b/w y-coordinates, how have you chosen the y coordinates? Cant it be -1-3=-4?[/quote]

Hi Harikamini,

Here two point are (0,-1) and (0,3).

Apply distance formula,

1) \(p=\sqrt{(3-(-1))^2+(0-0)^2}\)=4

2) \(p=\sqrt{(-1-3)^2+(0-0)^2}\)=4

Either way , distance is the same.

Hope it helps.
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Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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New post 12 Sep 2018, 06:33
fskilnik wrote:
[GMATH practice question]

Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two points which are p units apart. What is the value of p?

(A) 1/2
(B) 1
(C) 2
(D) 3
(E) 4


\(? = p\)

\(\left\{ \begin{gathered}
{\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} = 13 \hfill \\
x = 0 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\left( {y - 1} \right)^2} = 4\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
y - 1 = - 2 \hfill \\
\,\,\,\,\,{\text{OR}} \hfill \\
y - 1 = 2 \hfill \\
\end{gathered} \right.\)

\(\left\{ \begin{gathered}
x = 0\,\,\,;\,\,\,y - 1 = - 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{point}}\,\,\,A = \,\,\left( {0, - 1} \right) \hfill \\
\,\,\,\,\,{\text{OR}} \hfill \\
x = 0\,\,\,;\,\,\,y - 1 = 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{point}}\,\,B = \left( {0,3} \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\)

\(? = p = {\text{dist}}\left( {A,B} \right) = 4\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Our high-level "quant" preparation starts here: https://gmath.net
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Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin   [#permalink] 12 Sep 2018, 06:33
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