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

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 $$(1,-1)$$?

• you've nowhere used the other circle : $(x+2)^2 + (y-3)^2 = 25$, centered at $(-2,3)$ with a radius of $5$
– D S
Dec 28, 2022 at 7:06
• your circle just touches the other circle. We know that the line joining the two centres of a circle passes through their intersection point (or the mid-point of the intersection points, if there are 2), so the sum of distances from the centre of both circles to $(1,-1)$ is equal to the distance between the two centres
– D S
Dec 28, 2022 at 7:09
• How can that influence the answer? Where I am wrong? If it touches doesn't it mean that that point lies on it? Dec 28, 2022 at 7:10
• both the circles touch each other at $(1,-1)$ means that they only meet at $1$ point, which is a fundamental property
– D S
Dec 28, 2022 at 7:13
• So? how that solved the query? I am not getting it. Dec 28, 2022 at 7:15

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.

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 $$x^2+y^2-8x+10y=-16.$$

• yes thanks, I have corrected it now. Dec 30, 2022 at 5:56