If y = 3x + 1 is a tangent to x² + y² = k (k > 0 since it isa square), then where they meet has a repeated root; they meetat:x² + (3x + 1)² = k→ x² + 9x² + 6x + 1 - k = 0→ 10x² + 6x + (1 - k) = 0This is the point of contact when it has a repeated root whichis when the discriminant is zero, ie when:6² + 4 × 10 × (1 - k) = 0→ 36 + 40 - 40k = 0→ 40k = 4→ k = 1/10I guess for x & y you mean the point where y = 3x + 1 is atangent to x² + y² = k, ie the point of contact.The value of k can now be substituted into the equation of thepoint of contact:10x² + 6x + (1 - k) = 0→ 10x² + 6x + (1 - 1/10) = 0→ 10x² + 6x + 9/10 = 0→ x² + 6x/10 + 9/100 = 0→ (x + 3/10)²→ The point of contact is when x = -3/10→ y = 3× -3/10 + 1 = 9/10 + 1 = 1/10→ point of contact is (-3/10, 1/10) with k = 1/10