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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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11 Sep 2018, 06:45
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55% (hard)

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65% (02:08) correct 35% (02:02) wrong based on 36 sessions

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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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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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]

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11 Sep 2018, 07:01
2
3
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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WE: Supply Chain Management (Energy and Utilities)
Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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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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PKN

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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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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 [ 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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Joined: 11 Mar 2018
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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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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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Status: Learning stage
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WE: Supply Chain Management (Energy and Utilities)
Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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11 Sep 2018, 19:59
1
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.
_________________
Regards,

PKN

Rise above the storm, you will find the sunshine
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: Circle (x-3)^2 + (y-1)^2 = 13 intersects the vertical axis in two poin  [#permalink]

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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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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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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