Question

coordinates of smallest cicle touching x^2+y^2=4 and the line x+y=5 root2

Answer 1

Answer:

solved

Step-by-step explanation:

x+y=5√2 , its slope = -1 , slope of perpendicular = 1

equation of perpendicular passing through (5/√2 , 5/√2) is

y = x , substitute in  x^2+y^2=4 , y = x = 2/√2

coordinate of requred circle centre is (3.5/√2, 3.5/√2)

radius = 1.5

equation is

(x -  3.5/√2)² + (y - 3.5/√2)² = 1.5²