Question

12-38) the two circles x² + y² = 1 and (x - √2)² + (y - √2)² = 1 are tangent to each other. what are the coordinates of the point of tangency ?

Answer 1

`\left((√(2))/(2), (√(2))/(2) \right)`

Explanation:

We can solve the system.

`(x-√(2))^2+(y-√(2))^2=1 \implies (x^2+y^2)-2√(2)(x+y)+4=1 \\ \\ 1-2√(2)(x+y)+4=1 \\ \\ x+y=(1-1-4)/(-2√(2))=√(2) \implies y=√(2)-x \\ \\ \therefore x^2+(√(2)-x)^2=1 \\ \\ x^2+x^2-2√(2)x+2=1 \\ \\ 2x^2-2√(2)x+1=0 \\ \\ (√(2)x-1)^2=0 \\ \\ √(2)x=1 \\ \\ x=(√(2))/(2) \implies y=√(2)-(√(2))/(2)=(√(2))/(2) \\ \\ \therefore \left((√(2))/(2), (√(2))/(2) \right)`