Question

How do you write an equation of a circle whose center is at P(4,-5) and is tangent to the line -x+2y=1?

Answer 1

`x^2+y^2-8x+10y-4=0`

Explanation:

Since the line is tangent to the circle, its distance from the center is equal to the radius, that's unknown:

d=r

`d=(|ax_0+by_0+c|)/sqrt(a^2+b^2)`

where a,b,c are, in this case:

`a=-1; b=2;c=-1`

and `(x_0,y_0)` is the center `P(4,-5)`

So

radius=`|-1(4)+2(-5)-1|/sqrt((-1)^2+2^2)`

`=|-4-10-1|/sqrt5`

`=15/sqrt5=(15sqrt(5))/5=3sqrt(5)`

Now you can substitute the values in the equation of the circle:

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

`(x-4)^2+(y+5)^2=(3sqrt(5))^2`

`x^2+y^2-8x+10y+16+25-45=0`

`x^2+y^2-8x+10y-4=0`

Answer 2

`(x-4)^2+(y+5)^2- 45=0`

Explanation:

Giving a circle

`C->(x-x_0)^2+(y-y_0)^2-r^2=0`

and a line

`L->ax+by+c=0`

the tangency point `p_t={x_t,y_t}` is such that in `p_t` both declivities have the same value, or equivalently, their normal vectors are aligned.

Their normal vectors are

`vec n_C = {C_x,C_y} = {2(x_t-x_0),2(y_t-y_0)}` and
`vec n_L = {L_x,L_y} = {a,b}`

so at tangency

`vec n_C = lambda vec n_L`

The tangency point is obtained solving

#{ (2(x_t-x_0)+lambda a = 0), (2(y_t-y_0)+lambda b = 0), (a x_t + b y_t + c = 0) :}#

Here we know `x_0,y_0,a,b,c` and we are looking for `x_t,y_t,lambda`

but

#{ (x_t=-(a c - b^2 x_0 + a b y_0)/(a^2 + b^2)), (y_t = -(b c + a b x_0 - a^2 y_0)/(a^2 + b^2)), (lambda = (2 (c + a x_0 + b y_0))/(a^2 + b^2)) :}#

substituting the known values we obtain

`x_t=1,y_t=1,lambda=-6`

we know that

`r^2=(x_t-x_0)^2+(y_t-y_0)^2`

so the circle is

`C->(x-4)^2+(y+5)^2- 45=0`