Question

A circle passes through the point (-2,1) and touches the straight line 3x-2y=6 at the point (4,3).find its equation?

Answer 1

The equation is:
`(x+2/7)^2+(y-41/7)^2= ((10*sqrt(13))/7)^2`

Explanation:

From your equation: normal vector of p: `(3,-2)`.
You need to find point `S(AB)=(A+B)/2=((-2+4)/2,(3+1)/2)=(1,2)`.

Now you need to find the intersection of of k1 and k2. If (3,-2) is normal vec of p, than it is also directional vec of k1.

Similarly we can find vec of k2.
Direct. vec of AB is B-A -> (6,2). Normal vec of AB is than (2,-6) which is also direct. vec of k2.

Now we can build parametric equations of k1 and k2:
k1: `x=1+2t, y=2-6t` , (S(AB)=(1,2), k2 vec = (2,-6)).
k2: `x=4+3s, y=3-2s` , (B=(4,3), k1 vec = (3,-2)).

We put `x` from k1 equals `x` from k2 and the same for `y`.
We will get `s=-10/7` (or `t=-9/14`, one of them is enough in our case).
Fit `s` into parametric equation and you'll get:
`x=-2/7, y=41/7` which are coordinates of our beloved `S`!
`S=(-2/7,41/7)`.

The penultimate step: radius `r=|SA|` -> `SA=(-12/7,-34/7)`.
`|SA|=sqrt((-12/7)^2+(-34/7)^2)=(10*sqrt(13))/7=r`.

General circle equation: `(x-m)^2+(y-n)^2=r^2`.
After fitting: `(x-(-2/7))^2+(y-41/7)^2=((10*sqrt(13))/7)^2`.