Question

How do you write the equation of the circle passing through (−2,1) and tangent to the line 3x − 2y = 6 at the point (4, 3)?

Answer 1

`(x+2/7)^2+(y-41/7)^2=1300/49`

Explanation:

The circle equation is

`C->(x-x_0)^2+(y-y_0)^2= r^2`

Now it pass by the points `p_1=(-2,1)` and `p_2=(4,3)`

and also is tangent to the straight

`L->3x-2y-6=0`

So we have that for `C`

`2(x-x_0)dx+2(y-y_0)dy=0` or
`dy/dx=-(x-x_0)/(y-y_0)`

and also for `L`

`3dx-2dy=0` or
`dy/dx=3/2`

We also know that at `p_2`

`-(x_2-x_0)/(y_2-y_0)=3/2` so

`{((x_1-x_0)^2+(y_1-y_0)^2= r^2), ((x_2-x_0)^2+(y_2-y_0)^2=r^2), (-2(x_2-x_0)=3(y_2-y_0)):}`

Here the variables to find are `x_0,y_0,r`

Subtracting the first and the second we have a new reduced system

`{(x_1^2 - x_2^2 + y_1^2-y_2^2+2 x_0 (x_2-x_1) + 2 y_0( y_2 -y_1)=0),(-2(x_2-x_0)=3(y_2-y_0)):}` with solutions

`x_0=(3 x_1^2 - 3 x_2^2 + 3 (y_1 - y_2)^2 + 4 x_2 (y_2-y_1))/(6 (x_1 - x_2) - 4 y_1 + 4 y_2) = -2/7`

`y_0=((x_1 - x_2)^2 + y_1^2 + 3 (-x_1 + x_2) y_2 - y_2^2)/(3 (x_2 - x_1) + 2 (y_1 - y_2))=41/7`

finally

`r^2=(x_1-x_0)^2+(y_1-y_0)^2` giving

`r=(10 sqrt[13])/7`

Answer 2

Another approach to the problem is to consider that the circle's center `p_0` is located at the intersection of two lines: `p_0 = L_1 nn L_2` where

`L_1-> abs(p-p_1)^2=abs(p-p_2)^2`
`L_2->p = p_2+lambda vec v`

where `p = (x,y),p_1=(-2,1)` and `p_2=(4,3)`

From `L->3x-2y-6=0` we get `vec v=(2,3)`

the parameter `lambda in RR`

Expanding `L_1` we get

`L_1 -> < p,p_2-p_1 > =1/2(abs(p_2)^2-abs(p_1)^2)`

substituting `p` from `L_2` into `L_1` we get

`< p_2+lambda_0 vec v,p_2-p_1 > = abs(p_2)^2- < p_2, p_1 > + lambda < vec v, p_2-p_1 >` so

`lambda_0 = -(abs(p_2)^2+abs(p_1)^2+2 < p_2, p_1 >)/(< vec v, p_2-p_1>)`

then

`p_0=p_2+lambda_0 vec v` and also

`r = abs(p_0-p_2)`