Question

How do you write an equation for a circle whose center is at (0,0) and that is tangent to x+y=6?

Answer 1

`x^2+y^2=18`

Explanation:

The intersection of straight `s_1->y+x=6` and it's normal `s_2` passing by `p_c={x_c,y_c} = {0,0}` the circumference center, is also the tangency point between the circumference and the straight `s_1`.

The `s_2` equation can be obtained knowing that `m_1` is the `s_1` declivity. Then

`s_2->y = y_c -(1/m_1)(x-x_c)`

but `m_1 = ((dy)/(dx))_{s_1} = -1`

so

`s_2->y = 0 -(1/(-1))(x-0) = x`

`s_1 nn s_2->{(y +x = 6),(y - x = 0):}->{x_t=3,y_t=3}`

where `p_t = {x_t,y_t}` is the tangency point.

we know that `r = norm(p_c-p_t) = 3sqrt(2)`

The circumference equation reads

`(x-x_c)^2+ (y-y_c)^2=r^2`

or

`x^2+y^2=18`