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16 Sep 2014, 00:38
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Does the curve \((x - a)^2 + (y - b)^2 = 16\) intersect the \(y\) axis?
(1) \(a^2 + b^2 \gt 16\)
(2) \(a = |b| + 5\)
(1) \(a^2 + b^2 \gt 16\)
(2) \(a = |b| + 5\)
16 Sep 2014, 00:39
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Official Solution:
THEORY
In an xy coordinate system, the circle with center \((a, b)\) and radius \(r\) consists of all points \((x, y)\) that satisfy the following equation:
\((x-a)^2+(y-b)^2=r^2\)
This equation for the circle derives from the Pythagorean theorem applied to any point on the circle. As shown in the diagram above, the radius functions as the hypotenuse of a right-angled triangle whose other sides have lengths of x-a and y-b.
If the circle is centered at the origin (0, 0), the equation simplifies to: \(x^2+y^2=r^2\)
For more information, visit: Math: Coordinate Geometry
BACK TO THE ORIGINAL QUESTION
Does the curve \((x - a)^2 + (y - b)^2 = 16\) intersect the \(y\) axis?
The curve \((x - a)^2 + (y - b)^2 = 16\) is a circle with center at the point \((a, b)\) and a radius of \(\sqrt{16}=4\). Now, if \(a\), the x-coordinate of the center, is either greater than 4 or less than -4, then the radius of the circle, which is 4, would not be enough for the curve to intersect with the y-axis. Essentially, the question asks if \(|a| \gt 4\): if it is, then the answer will be NO, the curve does not intersect the y-axis. Conversely, if it's not, then the answer will be YES, the curve intersects with the y-axis.
(1) \(a^2 + b^2 \gt 16\).
This statement is clearly insufficient as \(|a|\) may or may not be more than 4.
(2) \(a = |b| + 5\).
Since the minimum value of the absolute value (in our case \(|b|\)) is zero, the minimum value of \('a\) would be 5. Therefore, in all cases, \(|a| \gt 4\), which means that the circle does not intersect the y-axis. Sufficient.
Answer: B
THEORY
In an xy coordinate system, the circle with center \((a, b)\) and radius \(r\) consists of all points \((x, y)\) that satisfy the following equation:
\((x-a)^2+(y-b)^2=r^2\)
This equation for the circle derives from the Pythagorean theorem applied to any point on the circle. As shown in the diagram above, the radius functions as the hypotenuse of a right-angled triangle whose other sides have lengths of x-a and y-b.
If the circle is centered at the origin (0, 0), the equation simplifies to: \(x^2+y^2=r^2\)
For more information, visit: Math: Coordinate Geometry
BACK TO THE ORIGINAL QUESTION
Does the curve \((x - a)^2 + (y - b)^2 = 16\) intersect the \(y\) axis?
The curve \((x - a)^2 + (y - b)^2 = 16\) is a circle with center at the point \((a, b)\) and a radius of \(\sqrt{16}=4\). Now, if \(a\), the x-coordinate of the center, is either greater than 4 or less than -4, then the radius of the circle, which is 4, would not be enough for the curve to intersect with the y-axis. Essentially, the question asks if \(|a| \gt 4\): if it is, then the answer will be NO, the curve does not intersect the y-axis. Conversely, if it's not, then the answer will be YES, the curve intersects with the y-axis.
(1) \(a^2 + b^2 \gt 16\).
This statement is clearly insufficient as \(|a|\) may or may not be more than 4.
(2) \(a = |b| + 5\).
Since the minimum value of the absolute value (in our case \(|b|\)) is zero, the minimum value of \('a\) would be 5. Therefore, in all cases, \(|a| \gt 4\), which means that the circle does not intersect the y-axis. Sufficient.
Answer: B
General Discussion
01 Oct 2014, 01:59
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Hi Bunuel I do not understand this "Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis" Can you explain it for me ? many thanks :D !
