Posted by **Lea** on Tuesday, January 29, 2013 at 10:34pm.

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

- Analytic Geometry -
**Reiny**, Tuesday, January 29, 2013 at 11:21pm
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

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