If the line x+ky=10 is tangent to the curve x^2 +y^2=10, find the possible values of k.?

Substituting x = 10 – ky into x^2 +y^2 – 10 = 0
(k^2 + 1)y^2 – 20ky + 90 = 0
Perfect square if 400k^2 = 360(k^2 + 1)
k = 3 or k = -3

Linex Of Ky

i) The line y = mx + c will be a tangent to the circle, if and only if its distance from the centre of the circle is equal to the radius of the circle.

ii) Applying this, the distance of the line x + ky - 10 = 0 from the centre (0, 0) is:
|(-10)/√((1² + k²)| = √10

Squaring, 100/(1 + k²) = 10
Solving k = ±3

ky = -x + 10
y = (-1/k) * x + 10

So it has a slope of (-1/k)

x^2 + y^2 = 10
2x * dx + 2y * dy = 0
y * dy = -x * dx
dy/dx = -x/y
dy/dx = -(10 - ky) / y

-1/k = -(10 - ky) / y
1/k = (10 - ky) / y
y = k * (10 - ky)
y = 10k - k^2 * y
y + y * k^2 = 10k
y * (1 + k^2) = 10k
y = 10k / (1 + k^2)

x + ky = 10
x + k * 10k / (1 + k^2) = 10
x + 10k^2 / (1 + k^2) = 10
x = 10 - 10k^2 / (1 + k^2)
x = (10 + 10k^2 - 10k^2) / (1 + k^2)
x = 10 / (1 + k^2)

y = 10k / (1 + k^2)
x = 10 / (1 + k^2)

x^2 + y^2 = 10
100 / (1 + k^2)^2 + 100k^2 / (1 + k^2)^2 = 10
(100 / (1 + k^2)^2) * (1 + k^2) = 10
100 / (1 + k^2) = 10
10 / (1 + k^2) = 1
10 = 1 + k^2
9 = k^2
-3 , 3 = k