Question

the point of contact of y=x+3√2 with x^2+y^2=9 is

Answer 1

Solution :

The given equations are

y = x + 3√2 .....(1)

x² + y² = 9 .....(2)

Using (1), from (2), we get

x² + (x + 3√2)² = 9

⇒ x² + x² + 6√2 x + 18 = 9

⇒ 2x² + 6√2 x + 9 = 0

⇒ x² + 3√2 x + 9/2 = 0

⇒ {x + (3√2)/2}² = 0

⇒ x + (3√2)/2 = 0

⇒ x = - (3√2)/2

⇒ x = - 3/√2

Putting x = - 3/√2 in (1) no. equation, we get

y = - 3/√2 + 3√2

⇒ y = (- 3 + 6)/√2

⇒ y = 3/√2

∴ the required point of intersection is (- 3/√2, 3/√2).