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# Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos

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GMATH Teacher
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Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

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14 Sep 2018, 08:48
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Difficulty:

45% (medium)

Question Stats:

63% (02:25) correct 37% (01:47) wrong based on 46 sessions

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[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7

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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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Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

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Updated on: 14 Sep 2018, 13:05
fskilnik wrote:
[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7

$$P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}$$

$${y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}$$

Let´s apply the "filling the squares" technique presented in our course!

$${x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\underbrace {{x^2} - 2x + \underline 1 }_{{{\left( {x - 1} \right)}^{\,2}}} + \underbrace {{y^2} - 4y + \underline 4 }_{{{\left( {y - 2} \right)}^{\,2}}} = \underbrace {4 + \underline 1 + \underline 4 }_9$$

$$P\,\, \in \,\,\,\left\{ {\,\,\left( {x,y} \right)\,\,:\,\,\,{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2} = {3^2}} \right\}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,P\,\, \in \,\,\,\, \odot \,\,\left\{ \begin{gathered} \,{\text{Centre}}\, = \left( {1,2} \right) \hfill \\ {\text{Radius}} = 3 \hfill \\ \end{gathered} \right.$$

$$\left. \begin{gathered} P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\, \in \,\,\,\, \odot \,\, \hfill \\ {y_P}\,\,\max \,\, \hfill \\ \end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{geometrically}}\,\,{\text{evident}}\,!} \,\,\,\,\,P = \left( {1,2 + 3} \right) = \left( {1,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 6$$

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

Originally posted by fskilnik on 14 Sep 2018, 08:49.
Last edited by fskilnik on 14 Sep 2018, 13:05, edited 1 time in total.
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Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

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14 Sep 2018, 09:22
1
fskilnik wrote:
[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7

We can write the equation as -
$$(x-1)^2 + (y-2)^2 = 9.$$ As y is max, by differentiating we get -
$$2(x-1) = 0$$ or $$x = 1, y =5$$

Another way -

As $$(x-1)^2 + (y-2)^2 = 9.$$
y is max when y = 5, x = 1

Hence, $$x + y = 6.$$
Option C.
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Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

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14 Sep 2018, 14:36
fskilnik wrote:
[GMATH practice question]

Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum possible vertical coordinate. What is the sum of the coordinates of P?

(A) 5
(B) 5.5
(C) 6
(D) 6.5
(E) 7

Alternate solution:

$$P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}$$

$${y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}$$

$${x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\, \Leftrightarrow \,\,\,\,{y^2} - 4y = 4 - \left( {{x^2} - 2x + \underline 1 } \right) + \underline 1 = 5 - {\left( {x - 1} \right)^2}$$

$${y^2} - 4y = 5 - {\left( {x - 1} \right)^2} \leqslant 5$$

$$y\,\,\max \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}} {x = {x_p} = 1} \\ {{y_p}^2 - 4{y_p} = 5} \end{array}\begin{array}{*{20}{c}} {} \\ {\,\,\,\mathop \Rightarrow \limits^{S = 4\,,\,P = - 5} \,\,\,\,{y_p} = \max \left\{ {5, - 1} \right\}\,\, = 5\,\,\,\,} \end{array}} \right.$$

$$? = 1 + 5 = 6$$

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

Regards,
Fabio.
_________________

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

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Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

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17 Sep 2018, 08:59
Can someone pls explain this question!
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 772
Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos  [#permalink]

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17 Sep 2018, 13:47
Ratnaa19 wrote:
Can someone pls explain this question!

Hi, Ratnaa19!

I suggest you try to understand my first solution, it is easier.

If you did not study the usual equation of a circle with given center and radius yet, I guess you should bookmark this problem for future reference.

On the other hand, if you know that equation and you want my help in any particular part of my solution, please ask in a very precise manner your doubt, so that I will be able to help you better.

Regards and success in your studies,
Fabio.
_________________

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

Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos   [#permalink] 17 Sep 2018, 13:47
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