Find the best way to solve this question

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June 21st 2009, 03:53 AM
simplependulum

Find the best way to solve this question

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

I've got the best way to solve this problem , do you know how i solved it ?
June 21st 2009, 04:32 AM
TheAbstractionist

Well, I did the usual textbook stuff

Spoiler:

First I found the equation of the line normal to $3x-y-1=0$ 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 $(4,1).$ Finally, I found the radius of the circle, which is $\sqrt{10}.$ Hence the circle is $(x-4)^2+(y-1)^2=10.$

Nevertheless the method did not appear to be particularly quick, so I guess you have a quicker solution.
June 21st 2009, 11:28 PM
simplependulum

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

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

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

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

Therefore , $k = -2$

Finally substitute back to F ,

The required equation is :

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

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