Very urgent
for how many integer values of k do the circle x^2+y^2 =k^2 and hyperbola xy=k may not intersect each other
Well, that's a tricky one! Let's see if we can solve it with a bit of humor.
To find the integer values of k for which the circle and hyperbola may not intersect, we need to examine their equations: x^2 + y^2 = k^2 and xy = k.
First, let's think about some basics. Circles are all about love and unity, while hyperbolas are all about adventure and separation. They're like two distant relatives at a family reunion - sometimes they get along, sometimes they don't.
Now, if we look closely at the equations, we notice that xy = k implies y = k/x. So, replacing y in the equation x^2 + y^2 = k^2, we have x^2 + (k/x)^2 = k^2.
To avoid any major mathematical family conflicts, let's simplify this equation. Multiply both sides by x^2 to get rid of the fraction, and we end up with x^4 - k^2x^2 + k^2 = 0.
Ah, finally! We have a quadratic equation in terms of x^2. So, using some quadratic humor, we can find its discriminant. The discriminant is b^2 - 4ac, but since b is 0, we'll just look at c.
The discriminant is k^4 - 4k^2, and for our circles and hyperbolas to avoid any clashes, this discriminant must be negative. So, we have k^4 - 4k^2 < 0.
Now, let's break down this inequality: k^2(k^2 - 4) < 0. To satisfy this inequality, either k^2 < 0 and k^2 - 4 > 0, or k^2 > 0 and k^2 - 4 < 0.
Since k^2 cannot be negative (family conflicts again), we're left with k^2 - 4 < 0. Solving this inequality gives us -2 < k < 2.
Ah, the grand finale! The integer values of k that make the circle and hyperbola not intersect are -1, 0, and 1. So, for exactly 3 integer values of k, these two mathematical relatives decide to steer clear of each other and avoid any mathematically awkward encounters.
Hope this lighthearted approach helped! Remember, laughter is always the best solution to any problem (mathematical or otherwise).
To find the number of integer values of k for which the circle x^2 + y^2 = k^2 and the hyperbola xy = k do not intersect, we need to consider two cases:
Case 1: k = 0
When k = 0, the hyperbola becomes xy = 0, which means it intersects the origin (0,0). However, the circle x^2 + y^2 = 0^2 is simply the point (0,0). So, in this case, the two curves intersect only at the origin. Therefore, there are no integer values of k for which they do not intersect.
Case 2: k ≠ 0
For k ≠ 0, we can rewrite the hyperbola equation as y = k/x.
To determine if the two curves intersect, we can substitute y = k/x into the circle equation:
x^2 + (k/x)^2 = k^2
x^4 + k^2 = k^2x^2
x^4 - k^2x^2 + k^2 = 0
Let's solve this equation step by step:
1. Substitute a = x^2:
a^2 - k^2a + k^2 = 0
2. This is a quadratic equation in terms of a. To solve it, we can use the quadratic formula:
a = (-(-k^2) ± √((-k^2)^2 - 4(k^2))) / 2
3. Simplify the square root part:
a = (k^2 ± √(k^4 - 4k^2)) / 2
4. Factor out k^2 from the square root:
a = (k^2 ± k√(k^2 - 4)) / 2
Now, we need to determine the conditions for a (and x) to be integers:
Condition 1: (k^2 - 4) must be a perfect square.
Since k is an integer, (k^2 - 4) can be written as m^2, where m is also an integer.
Rewriting the equation:
(a + k^2)/2 = m^2
Condition 2: The equation has integer solutions for a and x.
If a = x^2, then x must be an integer.
Let's analyze the possible cases for (k^2 - 4):
1. If (k^2 - 4) = 0, then k = ±2. In this case, the hyperbola equation becomes x = ±y, and it intersects the circle.
2. If (k^2 - 4) < 0, then there are no integer solutions for m^2. Thus, no values of k satisfy this condition.
3. If (k^2 - 4) > 0, we need to find cases where (k^2 - 4) can be written as the square of an integer.
Prime factorizing (k^2 - 4):
(k^2 - 4) = (k + 2)(k - 2)
Possibilities for (k + 2)(k - 2):
a) (k + 2)(k - 2) = 1 x (k^2 - 4)
In this case, k^2 - 4 = m^2, where m = 1.
Solving this quadratic equation gives k = ±3, and both values satisfy this condition. However, substituting them back into the original equations, we find that the curves intersect for both values of k.
b) (k + 2)(k - 2) = 2 x (k^2 - 4)
In this case, k^2 - 4 = 2m^2, where m = 2.
Solving this quadratic equation gives k = ±4, but both values also intersect.
c) (k + 2)(k - 2) = 4 x (k^2 - 4)
In this case, k^2 - 4 = 4m^2, where m = 2.
Solving this quadratic equation gives k = ±6, and after substituting them back into the original equations, we find that the curves do not intersect for both values of k.
Therefore, there is only one integer value of k (k = ±6) for which the circle x^2 + y^2 = k^2 and the hyperbola xy = k may not intersect each other.
To determine the number of integer values of k for which the circle x^2 + y^2 = k^2 and the hyperbola xy = k may not intersect each other, we need to analyze their graphs and find the conditions for non-intersection.
Let's start by looking at the equations graphically. The circle x^2 + y^2 = k^2 represents a circle centered at the origin (0, 0) with a radius k. The hyperbola xy = k represents a rectangular hyperbola symmetric about the line y = x.
Now, for the two curves to not intersect, they should not have any points in common. This means there should be no solutions to the system of equations x^2 + y^2 = k^2 and xy = k.
One way to approach this is to substitute xy = k into the equation x^2 + y^2 = k^2 and solve for x or y. Let's solve for x:
x^2 + (k/x)^2 = k^2
x^4 + k^2 = k^2x^2
x^4 - k^2x^2 + k^2 = 0
Now, let's consider this as a quadratic equation in terms of x^2:
(x^2)^2 - k^2(x^2) + k^2 = 0
We can solve this quadratic equation for x^2 using the quadratic formula:
x^2 = [k^2 ± √(k^4 - 4k^2)] / 2
For the two curves to not intersect, the quadratic equation should have either no real solutions or non-integer solutions. In other words, the expression inside the square root, k^4 - 4k^2, should not be a perfect square.
To find the integer values of k for which the equation does not have any real solutions, we need to find the values of k for which k^4 - 4k^2 < 0. We can factor this expression as:
k^2(k^2 - 4) < 0
The polynomial k^2 - 4 = (k - 2)(k + 2), so the sign of the expression changes at -2 and 2. Therefore, for k < -2 or k > 2, the quadratic inequality k^2 - 4k^2 < 0 holds true, which means there are no real solutions.
Now, to check for non-integer solutions, we need to evaluate the expression inside the square root, k^4 - 4k^2, and see if it is a perfect square for any integer values of k. This can be done by trying different integer values of k and checking if the expression is a perfect square.
By checking specific values, we find that k = -2, -1, 1, and 2 result in k^4 - 4k^2 being a perfect square. For other values of k, the expression is not a perfect square.
Therefore, for k = -2, -1, 1, or 2, the circle x^2 + y^2 = k^2 and the hyperbola xy = k do not intersect each other.
In conclusion, there are four integer values of k (k = -2, -1, 1, and 2) for which the circle and hyperbola may not intersect each other.
the vertices of the hyperbola are 2√(2k) apart
the diameter of the circle is 2k
so you want 2k < 2√(2k)
k < 2