# Thread: Find the best way to solve this question

1. ## Find the best way to solve this question

Find the equation of the circle which is tangent to the line $\displaystyle 3x - y - 1 = 0 ~$ at $\displaystyle A(1,2)~$ and passes through the point $\displaystyle B(5,-2)$

I've got the best way to solve this problem , do you know how i solved it ?

2. Well, I did the usual textbook stuff

Spoiler:
First I found the equation of the line normal to $\displaystyle 3x-y-1=0$ and passing through A. The centre of the circle lies on this line. Then I found the equation of the perpendicular bisector of the chord AB. The centre of the circle also lies on this line. Solving for the point of intersection, I determined the centre of the circle to be $\displaystyle (4,1).$ Finally, I found the radius of the circle, which is $\displaystyle \sqrt{10}.$ Hence the circle is $\displaystyle (x-4)^2+(y-1)^2=10.$

Nevertheless the method did not appear to be particularly quick, so I guess you have a quicker solution.

3. Let F be a family of circles such that all members of the family F touch the line $\displaystyle 3x - y - 1 = 0 ~~$at $\displaystyle A(1,2)$ .

Then F should be $\displaystyle (x-1)^2 + (y-2)^2 + k(3x - y - 1) = 0$ .

Since the required member passes through $\displaystyle B(5,-2)$ ,

it satisifies $\displaystyle (5-1)^2 + (-2-2)^2 + k[3(5) - (-2) - 1] = 0$ .

Therefore , $\displaystyle k = -2$

Finally substitute back to F ,

The required equation is :

$\displaystyle (x-1)^2 + (y-2)^2 -2(3x-y-1) = 0$
$\displaystyle x^2 + y^2 - 8x - 2y + 7 = 0$