Point P is the point of the circle x^2 + y^2 -2x -4y = 4 wit

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

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 :: www.GMATH.net (Math for the GMAT)
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Alternate solution (adapted from the very nice idea of regor60, posted above. Thank you for your contribution!):

\[P = \left( {{x_P}\,,\,\,{y_P}} \right)\,\,\, \in \,\,\,\,\left\{ {\,\left( {x,y} \right)\,\,\,:\,\,\,{x^2} - 2x + {y^2} - 4y = 4\,} \right\}\]
\[{y_P}\,\,\max \,\,\,,\,\,\,\,\,? = {x_P} + {y_P}\]
\[{x^2} - 2x + {y^2} - 4y = 4\,\,\,\,\, \Leftrightarrow \,\,\,\,{y^2} - 4y = 4 - \left( {{x^2} - 2x + \underline 1 } \right) + \underline 1 = 5 - {\left( {x - 1} \right)^2}\]
\[{y^2} - 4y = 5 - {\left( {x - 1} \right)^2} \leqslant 5\]
\[y\,\,\max \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}
{x = {x_p} = 1} \\
{{y_p}^2 - 4{y_p} = 5}
\end{array}\begin{array}{*{20}{c}}
{} \\
{\,\,\,\mathop \Rightarrow \limits^{S = 4\,,\,P = - 5} \,\,\,\,{y_p} = \max \left\{ {5, - 1} \right\}\,\, = 5\,\,\,\,}
\end{array}} \right.\]
\[? = 1 + 5 = 6\]


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

Regards,
Fabio.

_________________
Fabio Skilnik :: 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!

(x-h)² + (y-k)² = r² is a circle with a center at (h, k) and a radius of r.

x² + y² - 2x - 4y = 4

x² - 2x + y² - 4y = 4

x² - 2x + 1 + y² - 4y + 4 = 4 + 1 + 4

(x-1)² + (y-2)² = 9

(x-1)² + (y-2)² = 3³

The equation above constitutes a circle with a center at (1, 2) and a radius of 3.
Since r=3, the highest point is located exactly 3 units above the center:
As shown in the figure above, the highest point is at (1, 5).
Sum of the coordinates = 1+5 = 6.

The correct answer is C.

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