Find the value of k for which the line y = x+2 meets the curve y^2 + (x+k)^2 = 2?

Question

Find the value of k for which the line y = x+2 meets the curve y^2 + (x+k)^2 = 2

Answer 1

When the two curves meet, then their coordinates are the same. Thus being said, since the first curve y = x + 2, the equation of the second curve becomes

(x + 2)^2 + (x + k)^2 = 2

x^2 + 4x + 4 + x^2 + 2xk + k^2 = 2

2x^2 + 4x + 2xk + k^2 + 4 - 2 = 0

2x^2 + 4x + 2xk + k^2 + 2 = 0

k^2 + 2xk + 2(x^2 + 2x + 1) = 0

The above is a quadratic equation for "k" in terms of "x". Use the quadratic formula to determine the relationship of "k" and "x".

Answer 2

0=< k =<4

points of intersection are (-1,1) and (-3,1)

Anywhere between 0<k<4 has two intersection points.

It is a circle with radius √2 intersecting a line of 45 degrees.

Sketch it out to find the intersection points,

then solve it with 2 quadratic equations,

one for each point. You will get k.

For (-1,1), k is between (0,2)

For  (-3,1), k is between (2,4)

So k is between (0,4)

Here is the calculation: (in a minute)

For (-1,1), y² + (x+k)² = 2

1² + (-1+k)² = 2

k²-2k+2 = 2

k(k-2)=0

k = 0 and 2

The circle toughs the line when k =0

y²+x²=2 starts toughing the line at (-1,1).

As you move the circle to the left, the last

point of intersection is (-3,1).

For (-3,1) , y² + (x+k)² = 2

1² + (-3+k)² = 2

1 + k²-6k+9 = 2

k = 2 and 4

At k = 2, the intersections are the two points

At k = 4, the circle is at the furthest left toughing

only the point (-3,-1)

So k = 0 and 4 with only only one intersection.