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

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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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New post 14 Sep 2018, 09:48
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Question Stats:

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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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Course release PROMO : finish our test drive till 30/Sep with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 70% discount!

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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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New post Updated on: 14 Sep 2018, 14:05
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.
_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 30/Sep with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 70% discount!


Originally posted by fskilnik on 14 Sep 2018, 09:49.
Last edited by fskilnik on 14 Sep 2018, 14:05, 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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New post 14 Sep 2018, 10:22
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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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New post 14 Sep 2018, 15:36
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.
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Course release PROMO : finish our test drive till 30/Sep with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 70% discount!

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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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New post 17 Sep 2018, 09:59
Can someone pls explain this question!
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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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New post 17 Sep 2018, 14:47
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.

Thank you for your understanding.

Regards and success in your studies,
Fabio.
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Course release PROMO : finish our test drive till 30/Sep with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 70% discount!

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Re: Point P is the point satisfying x^2 + y^2 -2x -4y = 4 with maximum pos &nbs [#permalink] 17 Sep 2018, 14:47
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