The image of the centre of the circle x^(2)+y^(2)=a^(2) with respect to the mirror x+y=1 is
Answers
Answer:
(1 , 1)
Step-by-step explanation:
You can either use
2gx = h, 2fy = k, where (h, k) is the centre,
Comparing: 0 = h, 0 = k ∴ centre is (0, 0).
Or
directly, (x - 0)² + (y - 0)² = r², centre is (0, 0).
From all the cases, centre of the circle is (0, 0) and radius is 'a'.
Let image of (0, 0) on x + y = 1 be (α , β).
Line joining (0, 0) and (α, β) must be ⊥ to x + y = 1, so product of their slope is -1. [Slope of x + y = 1 is -1, on comparing with y = mx + c]
∴ (β - 0)/(α - 0) * (-1) = -1
∴ β = α
Since (0, 0) and (α, β) {now (α, α)} are mirror images on x + y = 1, there mid point must lie on the mirror i.e. x + y = 1.
Mid point of (0, 0) and (α, α) = (α/2 , α/2) [Mid point formula]
As (α/2 , α/2) lies on x + y = 1, it must satisfy x = α/2 and y = α/2
Therefore,
α/2 + α/2 = 1 ⇒ α = 1
hence, β = 1 [∵ β = α]
Therefore, image of the centre of the circle x^(2)+y^(2)=a^(2) is (1 , 1).
[Note that 'a' must have such value for which the circle doesn't intersects x + y =1 twice]