Question

for what values of k will the line be tangent to the circle ?x^2 + y^2 = 25 where y = x+k.please help I have an exam tomorrow

Answer 1

Explanation:

To find the values of k that will make the line y = x + k tangent to the circle x^2 + y^2 = 25, we need to set the equation of the line equal to the equation of the circle and solve for k.

First, we'll substitute y = x + k into the equation of the circle:

x^2 + (x + k)^2 = 25

Next, we'll expand the square on x + k:

x^2 + x^2 + 2xk + k^2 = 25

Now we'll combine like terms:

2x^2 + 2xk + k^2 = 25

We'll rearrange the equation so that x^2 is on one side

2x^2 + 2xk + k^2 - 25 = 0

Now we'll divide by 2

x^2 + xk + k^2/2 - 25/2 = 0

Now we'll complete the square

x^2 + xk + (k^2/2 - 25/2) = 0

x^2 + xk + (k^2/2 - 25/2) + (25/2) = (25/2)

(x + k/2)^2 = 25/2

Now we'll take the square root of both sides

x + k/2 = +-sqrt(25/2)

Now we'll solve for x

x = sqrt(25/2) - k/2 or x = -sqrt(25/2) - k/2

Now we'll substitute these values of x into the equation y = x + k and find the value of k.

x = sqrt(25/2) - k/2

y = x + k

y = sqrt(25/2) - k/2 + k

y = sqrt(25/2)

x = -sqrt(25/2) - k/2

y = x + k

y = -sqrt(25/2) - k/2 + k

y = -sqrt(25/2)

So the line y = x + k is tangent to the circle x^2 + y^2 = 25 when k = sqrt(25/2) or k = -sqrt(25/2)

Answer 2

Answer: To find the values of k for which the line y = x + k is tangent to the circle x^2 + y^2 = 25, we need to substitute y = x + k into the equation of the circle and solve for x.

x^2 + (x + k)^2 = 25

Expanding the square, we get:

x^2 + x^2 + 2xk + k^2 = 25

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

Now we can solve for x by using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Where a = 2, b = 2k and c = k^2

x = (-2k ± √(4k^2 - 42k^2)) / 2*2

x = (-k ± √(-3k^2)) / 4

Since the square root of a negative number is not a real number, we can see that there is no real value of x that will make the equation true. Therefore, the line y = x + k is never tangent to the circle x^2 + y^2 = 25