This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Substitute the equation of the line into the equation of the circle. The equation of the circle centred at M is (x+1)^2 + (y-1)^2 = 1. The equation of the straight line is y = -x + k. Substitute y = -x + k into the circle equation: (x+1)^2 + ((-x+k)-1)^2 = 1 (x+1)^2 + (-x + (k-1))^2 = 1 Step 2: Expand and simplify the equation. Expand the squared terms: (x^2 + 2x + 1) + ((-x)^2 + 2(-x)(k-1) + (k-1)^2) = 1 x^2 + 2x + 1 + x^2 - 2x(k-1) + (k^2 - 2k + 1) = 1 x^2 + 2x + 1 - 2kx + 2x + k^2 - 2k + 1 = 1 Combine like terms and move the constant to the left side to set the equation to 0: 2x^2 + (2 - 2k + 2)x + (1 + k^2 - 2k + 1 - 1) = 0 2x^2 + (4 - 2k)x + (k^2 - 2k + 1) = 0 This shows that the equation can be written as 2x^2 + (4 - 2k)x + (k^2 - 2k + 1) = 0. Step 3: Use the condition for tangency to find k. For the line to be a tangent to the circle, the quadratic equation 2x^2 + (4 - 2k)x + (k^2 - 2k + 1) = 0 must have exactly one real solution. This means its discriminant () must be equal to zero. The discriminant for a quadratic equation ax^2 + bx + c = 0 is = b^2 - 4ac. From our equation, we have: a = 2 b = (4 - 2k) c = (k^2 - 2k + 1) Set the discriminant to zero: = (4 - 2k)^2 - 4(2)(k^2 - 2k + 1) = 0 (16 - 16k + 4k^2) - 8(k^2 - 2k + 1) = 0 16 - 16k + 4k^2 - 8k^2 + 16k - 8 = 0 Combine like terms: (4k^2 - 8k^2) + (-16k + 16k) + (16 - 8) = 0 -4k^2 + 8 = 0 4k^2 = 8 k^2 = (8)/(4) k^2 = 2 k = ±sqrt(2) The numerical values of k are ±sqrt(2). 3 done, 2 left today. You're making progress.