What is the value of k when the straight line of y equals 3x plus 1 touches the curve of x squared plus y squared equals k at one point showing work with answer?

Answer 1

If: x^2+y^2 = k and y = 3x+1

Then: x^2+(3x+1)^2 = k => x^2+9x^2+6x+1 = k

Collecting like terms and subtracting k from both sides: 10x^2+6x+(1-k) = 0

For the line to touch the curve at one point the discriminant of b^2-4(ac) must = 0

So: 36-4*10*(1-k) = 0 => 36-40+40k = 0 => 40k = 40-36 => 40k = 4

Therefore: k = 4/40 and in its lowest terms 1/10