Find the equation of the circle with radius sqrt2 tangent to the line x+y=3 and having its center on the line y=4x

Answer 1

let the centre be (a,b)
but (a,b) lies on y = 4x
so b = 4a
and we can call our centre (a, 4a)
so the equation of our circle is

(x-a)^2 + (y-4a)^2 = 2

we know the distance from (a,4a) to
x+y-3 = 0 is √2

|a + 4a - 3|/√(1+1) = √2
5a - 3 = 2
5a = 5
a = 1

so the circle equation is
(x-1)^2 + (y-4)^2 = 2