# How do you find the equation for the circle radius 4 are tangent to the graph of y^2=4x at (1,2)?

Aug 11, 2016

See below

#### Explanation:

The circle equation is

$c \left(x , y\right) = {\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} - {r}^{2} = 0$

and the parabola

$p \left(x , y\right) = {y}^{2} - 4 x = 0$

At tangency point both curves have the same declivity or the same orientation in their normal vectors. Let ${p}_{t} = \left\{{x}_{t} , {y}_{t}\right\} = \left\{1 , 2\right\}$ their tangency point. At this point

$\nabla c \left({x}_{t} , {y}_{t}\right) + \lambda \nabla p \left({x}_{t} , {y}_{t}\right) = \vec{0}$

where

$\nabla c \left({x}_{t} , {y}_{t}\right) = \left\{2 \left({x}_{t} - {x}_{0}\right) , 2 \left({y}_{t} - {y}_{0}\right)\right\}$ and
$\nabla p \left({x}_{t} , {y}_{t}\right) = \left\{- 4 , 2 {y}_{t}\right\}$

and also

$c \left({x}_{t} , {y}_{t}\right) = {\left({x}_{t} - {x}_{0}\right)}^{2} + {\left({y}_{t} - {y}_{0}\right)}^{2} - {r}^{2} = 0$

Solving
{ (2(x_t-x_0) =-4 lambda), (2(y_t-x_0)=2 lambda y_t), ( (x_t-x_0)^2+(y_t-y_0)^2-r^2 = 0) :}

for ${x}_{0} , {y}_{0} , \lambda$ we obtain

( (x_0 = 1 - 2 sqrt[2], y_0 = 2 (1 + sqrt[2]), lambda = sqrt[2]), (x_0 = 1 + 2 sqrt[2], y_0 = 2 (1 - sqrt[2]), lambda = -sqrt[2]))

Attached a figure showing the two solutions.