Is the resultant, the locus of the center of the circle?

Question

If a circle $C$ passing through the point $(4,0)$ touches the circle $x^2 + y^2 + 4x − 6y = 12$ externally at the point $(1,−1)$, then the radius of $C$ is?

I have a question here, I have two points here on the circle $C$, $(4,0)$ and, $(1,-1)$ then the distance between the center point let's say $(x,y)$ and the two points will be equal.

So when I equate both of them I get this equation $6x+2y-14=0$ isn't this line the locus of the center of the circle?

If it is then why the radius is wrong when I used the distance of a point from a line, the line being $6x+2y-14=0$ and the point being $(1,-1)$?

Here is the link to this question: https://www.toppr.com/ask/question/if-a-circle-c-passing-through-the-point-4-0/

Answer 1

The eq. of tangent to $S=x^2+y^2+4x-6y-12=0$ at the point $(1,-1)$ is $L=x-y+2(x+1)-3(y-1)12=0\implies L=3x-4y-7=0$. Next the equation family of curves touching S and L externally is nothing but $$S'=S+tL= x^2+y^2+4x-6y-12+3tx-4ty-7t=0$$

Now let $S'$ pass through (4,0) to get the $t=-4$ to fix the required circle C as $x^2+y^2-8x+10y=-16.$

Answer 2

The given circle can be written

$$( x+2)^2+(y-3)^2=5^2 $$

that has center at $(-2,3)$ and radius $5.$

Finding the continuation line $CEG$ equation when RHS is evaluated at $ E (1,-1):$

$$\frac{y-3}{x+2}=\frac{-4}{3} \to 4 x +3 y= 1 \tag 1 $$

Next, you have already found the perpendicular bisector locus line as

$$3 x +y = 7 \tag 2$$

Solving equations 1,2 the center point coordinates are found as

$$ (x,y)= (4,-5) \tag 3 $$

It is given that the point passes through $$ (4,0) \tag 4 $$

The distance between above two points is easily found out, as the radius length, equal to 5.