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The equation of the circle can be written as
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18 Mar 2024, 05:49
18 Mar 2024, 05:49
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Question Stats:
75% (03:48) correct
25% (01:59) wrong based on 8 sessions
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The equation of the circle can be written as \((x-a)^2+(y-b)^2=r^2\) where (a,b) is the center of the circle and r is the radius.
What is the center point of the circle \(\frac{y+4}{y-4}=\frac{15}{(x-4)(y-2)}=\frac{x-2}{y-2}\)
A. (4,4)
B. (2,2)
C. (2,4)
D. (3,1)
E. ( 1,3)
What is the center point of the circle \(\frac{y+4}{y-4}=\frac{15}{(x-4)(y-2)}=\frac{x-2}{y-2}\)
A. (4,4)
B. (2,2)
C. (2,4)
D. (3,1)
E. ( 1,3)
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: The equation of the circle can be written as
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11 Jun 2024, 05:24
11 Jun 2024, 05:24
Expert Reply
This problem is not that difficult . However, its lengthy is troublesome.
To solve it faster you have two ways
1) Or manipulate the equation \(\frac{y+4}{y-4}=\frac{15}{(x-4)(y-2)}=\frac{x-2}{y-2}\) and essentially you will end up with the same equation \((x-a)^2+(y-b)^2=r^2\)
2) Or use the answer choices and see what is going on
In the first case after manipulation, we will have
\((x-3)^2+(y+1)^2=25\)
Where a=3, b=-1 and r=5
Choice D is correct
To solve it faster you have two ways
1) Or manipulate the equation \(\frac{y+4}{y-4}=\frac{15}{(x-4)(y-2)}=\frac{x-2}{y-2}\) and essentially you will end up with the same equation \((x-a)^2+(y-b)^2=r^2\)
2) Or use the answer choices and see what is going on
In the first case after manipulation, we will have
\((x-3)^2+(y+1)^2=25\)
Where a=3, b=-1 and r=5
Choice D is correct
Re: The equation of the circle can be written as
[#permalink]
18 Jun 2024, 05:02
18 Jun 2024, 05:02
1
Combining y+4/y-4 = x-2/y-2
y^2 -2y +4y -8 = (y-4)(x-2)
y^2 +2y -8 = (y-4)(x-2)...............1
15/(x-4)(y-2) = x-2/y-2
15 = x^2 -6x+ 8 =15
x^2 -6x -7=0.................2
Sum of the two equations =0 so
x^2 -6x -7 + y^2 + 2y -8 = 0
x^2 -6x + y^2 +2y = 15
x^2 -6x +9 + y^2 +2y + 1 = 15 + 10
(x-3)^2 + (y + 1) ^2 = 5^2
So coordinate of centre = ( 3, -1) no such answer in the option.
Adewale Fasipe, GRE, GMAT Quant instructor from Lagos Nigeria.
y^2 -2y +4y -8 = (y-4)(x-2)
y^2 +2y -8 = (y-4)(x-2)...............1
15/(x-4)(y-2) = x-2/y-2
15 = x^2 -6x+ 8 =15
x^2 -6x -7=0.................2
Sum of the two equations =0 so
x^2 -6x -7 + y^2 + 2y -8 = 0
x^2 -6x + y^2 +2y = 15
x^2 -6x +9 + y^2 +2y + 1 = 15 + 10
(x-3)^2 + (y + 1) ^2 = 5^2
So coordinate of centre = ( 3, -1) no such answer in the option.
Adewale Fasipe, GRE, GMAT Quant instructor from Lagos Nigeria.
