Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

THEORY
In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)




This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\)

For more check: math-coordinate-geometry-87652.html

BACK TO THE ORIGINAL QUESTION

Does the curve \((x - a)^2 + (y - b)^2 = 16\) intersect the \(Y\) axis?

Curve of \((x - a)^2 + (y - b)^2 = 16\) is a circle centered at the point \((a, \ b)\) and has a radius of \(\sqrt{16}=4\). 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. So basically the question asks whether \(|a|>4\): if it is, then the answer will be NO: the curve does not intersect with Y axis and if it's not, then the answer will be YES: the curve intersects with Y axis.

(1) \(a^2 + b^2 > 16\) --> clearly insufficient as \(|a|\) may or may not be more than 4.

(2) \(a = |b| + 5\) --> as the least value of absolute value (in our case \(|b|\)) is zero then the least value of \(a\) will be 5, so in any case \(|a|>4\), which means that the circle does not intersect the Y axis. Sufficient.

Answer: B.