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

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Fri Sep 14, 2018 8:42 am

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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

Answer: [spoiler]__(C)____[/spoiler]

Last edited by fskilnik@GMATH on Fri Sep 14, 2018 8:50 am, edited 1 time in total.

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Source: Beat The GMAT — Problem Solving | Fri Sep 14, 2018 8:42 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

by fskilnik@GMATH » Fri Sep 14, 2018 8:44 am

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

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

LetÂ´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.

Last edited by fskilnik@GMATH on Sat Sep 15, 2018 4:00 pm, edited 2 times in total.

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

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

by regor60 » Fri Sep 14, 2018 9:51 am

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

Answer: [spoiler]__(C)____[/spoiler]

Reorganizing the problem: y^2 - 4y +x^2- 2x = 4

To maximize the vertical is to maximize y. To maximize y, we need to minimize the expression x^2 - 2x since to the extent it is positive will reduce the right side, 4, available for the y expression to satisfy.

Looking at x^2 - 2x it is tempting to believe the minimum value is 0, but can we do better than that ?

If we set x=1, then the expression reduces to -1, that's better than 0. If we set x=2, the expression = 0, so headed in the wrong direction.

If we set x = 0, the expression = 0 , again the wrong direction. Testing x=-1 the expression = 3.

So x = 1 minimizes the expression. Plugging x=1 into the expression and solving for y:

y^2-4y + 1 - 2 = 4 > y^2-4y - 5 =0

Solving for Y: (y+1)(y-5) = 0 so y = -1, 5. The maximum is 5.

So the coordinates of the point P where y is a maximum, 5, are (1,5) so the sum of the coordinates is 6, C

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

Point P is the point on the circle x^2 + y^2 -2x -4y = 4 wi

by fskilnik@GMATH » Fri Sep 14, 2018 12:30 pm

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 (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.

Last edited by fskilnik@GMATH on Fri Sep 14, 2018 12:58 pm, edited 1 time in total.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Sep 14, 2018 12:36 pm

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

(x-h)Â² + (y-k)Â² = rÂ² is a circle with a center at (h, k) and a radius of r.

xÂ² + yÂ² - 2x - 4y = 4

xÂ² - 2x + yÂ² - 4y = 4

xÂ² - 2x + 1 + yÂ² - 4y + 4 = 4 + 1 + 4

(x-1)Â² + (y-2)Â² = 9

(x-1)Â² + (y-2)Â² = 3Â³

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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