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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Point P is the point of the circle x^2 + y^2 -2x -4y = 4 wit tagged by: fskilnik@GMATH ##### This topic has 3 expert replies and 1 member reply ### GMAT/MBA Expert ## Point P is the point of the circle x^2 + y^2 -2x -4y = 4 wit ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [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: __(C)____ _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br Last edited by fskilnik@GMATH on Fri Sep 14, 2018 8:50 am; edited 1 time in total ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 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 :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br Last edited by fskilnik@GMATH on Sat Sep 15, 2018 4:00 pm; edited 2 times in total ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 329 messages Upvotes: 27 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: __(C)____ 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 ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 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. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br Last edited by fskilnik@GMATH on Fri Sep 14, 2018 12:58 pm; edited 1 time in total ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15313 messages Followed by: 1863 members Upvotes: 13060 GMAT Score: 790 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)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. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! 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