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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/

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    $\begingroup$ you've nowhere used the other circle : $(x+2)^2 + (y-3)^2 = 25$, centered at $(-2,3)$ with a radius of $5$ $\endgroup$
    – D S
    Dec 28, 2022 at 7:06
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    $\begingroup$ 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 $\endgroup$
    – D S
    Dec 28, 2022 at 7:09
  • $\begingroup$ How can that influence the answer? Where I am wrong? If it touches doesn't it mean that that point lies on it? $\endgroup$
    – Ankit
    Dec 28, 2022 at 7:10
  • $\begingroup$ both the circles touch each other at $(1,-1)$ means that they only meet at $1$ point, which is a fundamental property $\endgroup$
    – D S
    Dec 28, 2022 at 7:13
  • $\begingroup$ So? how that solved the query? I am not getting it. $\endgroup$
    – Ankit
    Dec 28, 2022 at 7:15

2 Answers 2

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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 $$ (4,0) \tag 4 $$

enter image description here

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

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

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  • $\begingroup$ yes thanks, I have corrected it now. $\endgroup$
    – Z Ahmed
    Dec 30, 2022 at 5:56

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