# This question is really hard! i dont even understand. ?

### 5 Answers

- Anonymous1 month agoFavourite answer
y= x + 2 is a straight line. and y² + (x +k) ² = 2 is a circle.

Possibilities are, the straight line can:

a) not intersect the circle

b) touch the circle at 1 point (so the line is a tangent)

c) cut the circle at 2 points.

To find point(s) of intersection we put y = x + 2 into the circle’s equation:

(x+2)² + (x +k) ² = 2

x² + 4x + 4 + x² + 2kx + k² = 2

2x² + (2k+4)x + (k² + 2) = 0

This is a quadratic equation. For the straight line to be a tangent, it must only touch at one point so we need to find the value of k giving only one solution – so the discriminant of the quadratic equation (b² – 4ac) must be zero:

(2k+4)² - 4(2)(k² + 2) = 0

4k² + 16k + 16 – 8k² – 16 = 0

4k² + 16k - 8k² = 0

k² – 4k = 0

k(k – 4) = 0

k = 0 or k = 4

- ?Lv 71 month ago
Even though it's presented as a variable here, k will be a constant when graphing y^2 + (x+k)^2 = 2 on the y vs. x plane.

So the curve is a circle.Take the derivative to find the slope of the circle at any point (x,y).Use that to write the equation of a tangent line through any point (a, ±√(2 - (a+k)^2)) on the circle.You want that line to be the same as y = x + 2. That means all the coefficients have to be equal.For example, if you got that the tangent is y = (2a + 3k)x + 4a, then (2a+3k) must equal 1 and 4a must equal 2.Write out this set of equations. Solve for (a and) k.

hint: one solution is k = 0.

~~~~~~~~~~~~~~~~~~~~~~~~~

We can instead approach the problem with geometry instead of calculus.

y^2 + (x+k)^2 = 2 is a circle with radius √2 and center (-k, 0).

Every tangent line to a circle is perpendicular to a radius. Thus, where the tangent has slope 1, the radius has slope -1.

From a point (a, a+2) on the line, the center of the circle must be distance √2 along a -1 slope. What are the coordinates where the center could be?

Given that the center is (-k, 0), solve for k.

- Ian HLv 71 month ago
y^2 + x^2 = 2 is a circle centred (0, 0)

y^2 + (x + k)^2 = 2 is a similar circle centred (-k, 0)

Substitute y = x + 2 to investigate intersection of line with circle

(x + 2)^2 + (x + k)^2 – 2 = 0

2x^2 + (2k + 4)x + k^2 + 2 = 0 and for “b^2 = 4ac” as tangent

(2k + 4)^2 = 8(k^2 + 2)

0 = 4k^2 – 16k = 4k(k – 4)

k = 0 or 4

- Anonymous1 month ago
For any specific value of k, the curve in the second equation is a circle of radius √2, centered at (-k, 0). Change the value of k and the curve is still a circle and the radius is still √2. It's just centered at a new point; and that point is always on the x-axis.

There's no k in the first equation, so the line does not move. It's a 45-degree line through points (-2, 0) and (0, 2).

If this was a calculus problem, those observations makes it a geometry problem. Large positive values of k put the circle completely to the left of the line, while large negative values of k put it completely to the right. Where can you put the circle so it just touches the line?

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- az_lenderLv 71 month ago
The curve y^2 + (x + k)^2 = 2 is a circle of radius sqrt(2), and its center is at (k,0). Changing the value of k moves the center of the circle left and right along the x axis. There will be two positions where the circle will just touch the diagonal line y = x + 2 without crossing it.

The slope of the tangent is 1, so the slope of the circle's radius at the point of tangency will be -1. The radius of the circle is sqrt(2), and that is the same as the distance from the point of tangency back to the x-axis, measured along the line y = x + 2. So the right triangle formed by the circle's radius, the line y = x+2, and the x-axis has a hypotenuse of 2.

Therefore the center of the circle is 2 units away from the x-intercept of the line y = x + 2. The center of the circle is either at x = -4 or at x = 0. The value of "k" in the problem statement must be either 0 or 4.