# Two circles of radius `3*sqrt 2` are tangent to the graph y^2 = 4x at point (1, 2). Find the equation of these two circles.

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### 1 Answer

Two circles of radius `3*sqrt 2` are tangent to the graph y^2 = 4x at point (1, 2).

Two circles with radius 3*sqrt are tangent to the curve y^2 = 4x at the point (1, 2).

The tangent to the curve y^2 = 4x at (1, 2) is perpendicular to the line joining the center of the two circles and bisects it.

The tangent to y^2 = 4x at (1,2) has a slope `dy/dx` = `(4/(2y)) = 4/4 = 1` . The equation of the tangent is y = x + 1

The line joining the centers of the circle has a slope of -1. The equation of the line is `(y - 2)/(x - 1) = -1`

=> y - 2 = 1 - x

=> y = 3 - x

A point that lies on this line is of the form (x, -x + 3)

The distance between (x, -x+3) and (1, 2) is `3*sqrt 2`

=> (x - 1)^2 + (3 - x - 2)^2 = 12

=> x^2 -2x + 1 + 1 + x^2 - 2x = 12

=> x^2 - 2x + 1 = 6

=> x^2 - 2x - 5 = 0

x1 = `1 - sqrt 6`

x2 = `1 + sqrt 6`

y1 = `2 + sqrt 6`

y2 = `2 - sqrt 6`

The centers of the required circles are `(1 - sqrt 6, 2 + sqrt 6)` and `(1 + sqrt 6, 2 - sqrt 6)`

**The equation of the circles are `(x - 1 + sqrt 6)^2 + (y - 2 - sqrt 6)^2 = (3*sqrt 2)^2` and `(x - 1 - sqrt 6)^2 + (y - 2 + sqrt 6)^2 = (3*sqrt 2)^2` **