Q.126. Circle,Parabola, Tangent

Question 126.
Find the equations of two circles of radius 4 and tangent to the parabola y^2 = 4x at (1,2).

Answer 126.
Equation of tangent to y^2 = 4x at (1, 2) is
2y = 2(x + 1)
=> x - y + 1 = 0

Line perpendicular to it has the form
x + y = k
If it passes through (1, 2), its equation is
x + y = 3

The centers of the two circles are on x + y = 3 at a distance = 4 from (1, 2)
Slope of x + y = 3 is m = - 1
=> cosθ = - 1/√2 and sinθ = 1/√2
=> Centers of the two circles are
[1 + 4cosθ, 2 + 4sinθ] = (1 - 2√2, 2 + 2√2) and
[1 - 4cosθ, 2 - 4sinθ] = (1 + 2√2, 2 - 2√2)

 => equations of the required circles are
(x - 1 + 2√2)^2 + (y - 2 - 2√2)^2 = 4^2 and
(x - 1 - 2√2)^2 + (y - 2 + 2√2)^2 = 4^2
i.e.,
x^2 + y^2 - 2(1 - 2√2)x - 4(1 + √2)y + 5 + 4√2 = 0 and
x^2 + y^2 - 2(1 + 2√2)x - 4(1 - √2)y + 5 - 4√2 = 0

Refer to the plots of the given parabola and the derived equations of the circles in
Wolfram/Alpha link

LINK to YA!