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In the xycoordinate system, a circle with radius root30 and center
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Updated on: 24 Jan 2018, 23:40
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In the xycoordinate system, a circle with radius \(\sqrt {30}\) and center (2,1) intersects the xaxis at (k,0). One possible value of k is. A. \(2 + \sqrt {26}\) B. \(2 + \sqrt {29}\) C. \(2 + \sqrt {31}\) D. \(2 + \sqrt {34}\) E. \(2 + \sqrt {35}\)
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Originally posted by PathFinder007 on 08 Aug 2015, 11:48.
Last edited by Bunuel on 24 Jan 2018, 23:40, edited 3 times in total.
Renamed the topic and edited the question.



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Re: In the xycoordinate system, a circle with radius root30 and center
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16 Aug 2015, 16:53
PathFinder007 wrote: In the xycoordinate system, a circle with radius \(\sqrt30\) and center (2,1) intersects the xaxis at (k,0). One possible value of k is.
A. 2 + \(\sqrt 26\) B. 2 +\(\sqrt 29\) C. 2+ \(\sqrt 31\) D. 2 + \(\sqrt 34\) E. 2 +\(\sqrt 35\) Equation of circle in xy coordinates with center at (a,b) and radius r > \((xa)^2+(yb)^2 = r^2\) Per the question, a=2, b=1, \(r^2\)=30 We get the equation of the circle as \(x^2+y^24x2y = 25\) Now the above equation should also be satisfied by the point (k,0) such that we get \(k^24k25 = 0\) > k = 2+ \(\sqrt{29}\) as 1 possible value. B is the correct answer.



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Re: In the xycoordinate system, a circle with radius root30 and center
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17 Aug 2015, 08:42
Engr2012 wrote: PathFinder007 wrote: In the xycoordinate system, a circle with radius \(\sqrt30\) and center (2,1) intersects the xaxis at (k,0). One possible value of k is.
A. 2 + \(\sqrt 26\) B. 2 +\(\sqrt 29\) C. 2+ \(\sqrt 31\) D. 2 + \(\sqrt 34\) E. 2 +\(\sqrt 35\) Equation of circle in xy coordinates with center at (a,b) and radius r > \((xa)^2+(yb)^2 = r^2\) Per the question, a=2, b=1, \(r^2\)=30 We get the equation of the circle as \(x^2+y^24x2y = 25\)Now the above equation should also be satisfied by the point (k,0) such that we get \(k^24k25 = 0\) > k = 2+ \(\sqrt{29}\) as 1 possible value. B is the correct answer. How are you getting the highlighted portion as = 25? shouldn't it be 29? And in the last step, how did you eliminate the K from 4K?



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Re: In the xycoordinate system, a circle with radius root30 and center
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17 Aug 2015, 08:52
nitika231 wrote: Engr2012 wrote: PathFinder007 wrote: In the xycoordinate system, a circle with radius \(\sqrt30\) and center (2,1) intersects the xaxis at (k,0). One possible value of k is.
A. 2 + \(\sqrt 26\) B. 2 +\(\sqrt 29\) C. 2+ \(\sqrt 31\) D. 2 + \(\sqrt 34\) E. 2 +\(\sqrt 35\) Equation of circle in xy coordinates with center at (a,b) and radius r > \((xa)^2+(yb)^2 = r^2\) Per the question, a=2, b=1, \(r^2\)=30 We get the equation of the circle as \(x^2+y^24x2y = 25\)Now the above equation should also be satisfied by the point (k,0) such that we get \(k^24k25 = 0\) > k = 2+ \(\sqrt{29}\) as 1 possible value. B is the correct answer. How are you getting the highlighted portion as = 25? shouldn't it be 29? And in the last step, how did you eliminate the K from 4K? No, check your solution. When we get, \((xa)^2+(yb)^2 = r^2\) with a =2, b =1 and \(r^2\)=30, we get \((x2)^2+(y1)^2 = 30\) > \(x^2+44x+y^2+12y = 30\)> \(x^24x+y^22y+5 = 30\) > \(x^24x+y^22y = 25\). Thus, based on my post above, we get \(k^24k25 = 0\) We know that for any quadratic equation, \(px^2+qx+r =0\) with p,q,r as constants, the solutions are : \(x = \frac{q \pm \sqrt{q^24pr}}{2p}\), where p=1, q =4, r=25 Substituting the above values, we get, k = 2 +\(\sqrt 29\) Hope this helps.



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Re: In the xycoordinate system, a circle with radius root30 and center
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17 Aug 2015, 23:52
Quote: No, check your solution.
When we get, \((xa)^2+(yb)^2 = r^2\) with a =2, b =1 and \(r^2\)=30, we get
\((x2)^2+(y1)^2 = 30\) > \(x^2+44x+y^2+12y = 30\)> \(x^24x+y^22y+5 = 30\) > \(x^24x+y^22y = 25\).
Thus, based on my post above, we get
\(k^24k25 = 0\)
We know that for any quadratic equation, \(px^2+qx+r =0\) with p,q,r as constants, the solutions are : \(x = \frac{q \pm \sqrt{q^24pr}}{2p}\), where p=1, q =4, r=25
Substituting the above values, we get, k = 2 +\(\sqrt 29\)
Hope this helps. Understood! Thanks a lot for explaining it stepbystep! Maths has never been my strong suit, therefore, I struggle with what others may think is common knowledge. I realized I had miscalculated in one step and forgotten entirely about that quadratic formula in the second step.



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Re: In the xycoordinate system, a circle with radius root30 and center
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18 Aug 2015, 02:19
Quote: In the xycoordinate system, a circle with radius 3‾√0 and center (2,1) intersects the xaxis at (k,0). One possible value of k is.
A. 2 + 2‾√6 B. 2 +2‾√9 C. 2+ 3‾√1 D. 2 + 3‾√4 E. 2 +3‾√5 So we know the equation of the circle is \((xh)^2 + (yk)^2 = r^2\). Where (h,k) is the centre of the circle and r is the radius. In your question, we know (h,k) = (2,1) and r = \(\sqrt{30}\) Hence we get \((xh)^2 + (yk)^2 = r^2 as, (x2)^2 + (y1)^2 = 30\) now the point (k,0) should satisfy this circle's equation. So lets replace x and y with k and 0 respectively. \((k2)^2 + (01)^2 = 30\) =>\(k^2  4k = 25\) =>\(k^2  4k  25 =0\) => \(2 + \sqrt{29}\)
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Re: In the xycoordinate system, a circle with radius root30 and center
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10 Nov 2016, 05:44
Since the center is (2,1), drawing a perpendicular line to xaxis gives us a value (2,0). Given radius as root 30 & the point intersects xaxis at (k,0), therefore as per hypotenuse formula: 1^2+x^2=(root 30)^2. Solving we get x=root 29. Points of intersection is (root 29,0) & (2+root 29,0) Answer choice: B



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In the xycoordinate system, a circle with radius root30 and center
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22 Sep 2018, 11:18
Mistral wrote: Quote: In the xycoordinate system, a circle with radius 3‾√0 and center (2,1) intersects the xaxis at (k,0). One possible value of k is.
A. 2 + 2‾√6 B. 2 +2‾√9 C. 2+ 3‾√1 D. 2 + 3‾√4 E. 2 +3‾√5 So we know the equation of the circle is \((xh)^2 + (yk)^2 = r^2\). Where (h,k) is the centre of the circle and r is the radius. In your question, we know (h,k) = (2,1) and r = \(\sqrt{30}\) Hence we get \((xh)^2 + (yk)^2 = r^2 as, (x2)^2 + (y1)^2 = 30\) now the point (k,0) should satisfy this circle's equation. So lets replace x and y with k and 0 respectively. \((k2)^2 + (01)^2 = 30\) =>\(k^2  4k = 25\) =>\(k^2  4k  25 =0\) => \(2 + \sqrt{29}\) hey pushpitkc i encountered algebraic tech issues how to solve it through factoring out \(k^2  4k  25 =0\) to get 25 the only numbers that get multiplied are 5*5 or 1*25 , which dont give me 4 when adding up Hey niks18 maybe you can help ? pushpitkc is travelling I think So what am I missing



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In the xycoordinate system, a circle with radius root30 and center
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23 Sep 2018, 02:29
dave13 wrote: Mistral wrote: Quote: In the xycoordinate system, a circle with radius 3‾√0 and center (2,1) intersects the xaxis at (k,0). One possible value of k is.
A. 2 + 2‾√6 B. 2 +2‾√9 C. 2+ 3‾√1 D. 2 + 3‾√4 E. 2 +3‾√5 So we know the equation of the circle is \((xh)^2 + (yk)^2 = r^2\). Where (h,k) is the centre of the circle and r is the radius. In your question, we know (h,k) = (2,1) and r = \(\sqrt{30}\) Hence we get \((xh)^2 + (yk)^2 = r^2 as, (x2)^2 + (y1)^2 = 30\) now the point (k,0) should satisfy this circle's equation. So lets replace x and y with k and 0 respectively. \((k2)^2 + (01)^2 = 30\) =>\(k^2  4k = 25\) =>\(k^2  4k  25 =0\) => \(2 + \sqrt{29}\) hey pushpitkc i encountered algebraic tech issues how to solve it through factoring out \(k^2  4k  25 =0\) to get 25 the only numbers that get multiplied are 5*5 or 1*25 , which dont give me 4 when adding up Hey niks18 maybe you can help ? pushpitkc is travelling I think So what am I missing Hi dave13While factorization works well for most GMAT like questions but you should also remember the formula to calculate roots of the equation which you can use for complex quadratic equation such as the above. The formula to calculate the roots of the quadratic equation is \(b±\sqrt{b^24ac}/2a\) substitute the values of a,b & c in this formula, you will get your answer



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In the xycoordinate system, a circle with radius root30 and center
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23 Sep 2018, 02:35
Quote: Hi dave13While factorization works well for most GMAT like questions but you should also remember the formula to calculate roots of the equation which you can use for complex quadratic equation such as the above. The formula to calculate the roots of the quadratic equation is \(b±\sqrt{b^24ac}/2a\) substitute the values of a,b & c in this formula, you will get your answer Hi niks18 thank you for the explanaton, i solved it through the formula that you provided yesterday , but I couldnt solve it through factorization. can you please explain how to solve this \(k^2  4k  25 =0\) through factorozation as i mentioned earlier, to get 25 the only numbers that get multiplied are 5*5 or 1*25 , which dont give me 4 when adding up that is a stumbilng stone for me so what am i doing wrong ? thanks! have a great weekend



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Re: In the xycoordinate system, a circle with radius root30 and center
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23 Sep 2018, 03:07
dave13 wrote: Quote: Hi dave13While factorization works well for most GMAT like questions but you should also remember the formula to calculate roots of the equation which you can use for complex quadratic equation such as the above. The formula to calculate the roots of the quadratic equation is \(b±\sqrt{b^24ac}/2a\) substitute the values of a,b & c in this formula, you will get your answer Hi niks18 thank you for the explanaton, i solved it through the formula that you provided yesterday , but I couldnt solve it through factorization. can you please explain how to solve this \(k^2  4k  25 =0\) through factorozation as i mentioned earlier, to get 25 the only numbers that get multiplied are 5*5 or 1*25 , which dont give me 4 when adding up that is a stumbilng stone for me so what am i doing wrong ? thanks! have a great weekend Hi dave13Not all quadratic equation can be solved using factorization method. Hence there are multiple methods. this is one of the equation that cannot be solved using factorization. consider another example \(x^23x3=0\). this equation can also be not solved using factorization. You need to use other methods depending upon the equation.




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