Doubt in why a method of coordinate geometry worked. My professor had given us a problem. It goes like -

Find equation of circle passing through intersection of circle $S=x^2+y^2-12x-4y-10=0$ and $L:3x+y=10$ and having radius equal to that of circle $S$.

So when he was telling the solution, he setup an equation like -
$x^2+y^2-12x-4y-10=0+2t(3x+y-10)=0$ where $t \in \mathbb{R}$
$= x^2+y^2+2(3t-6)x+2(-2+t)y-(10+2t)=0$
So now we can find out that radius of $S=\sqrt{50}$
Now we can use for any general circle $r=\sqrt{g^2+f^2-c}$ to get ,
$(6-3t)^2+(t-2)^2+10+20t=50\implies t=0,2$
Now we can remove $t=0$ as we will get $S$
So putting $t=2 \implies \boxed{x^2+y^2=50}$
Now this solution is ok, but I didn't get why putting $t$ worked? I mean why is $x^2+y^2-12x-4y-10=0+2t(3x+y-10)=0$ working? Can someone explain this?
 A: Let $A,B$ be the intersection points of $S$ and $L$; in other words $S(A)=S(B)=L(A)=L(B)=0$. Then for any $t$ we also have $(S+tL)(A)=(S+tL)(B)=0$, that is $S+tL$ passes through $A$ and $B$. You seem to be asking why these are all the circles that pass through $A$ and $B$. I think this is usually explained via linear algebra. Very roughly speaking, a general circle looks like $x^2+y^2+ax+by+c=0$, with $3$ coefficients, so "the degree of freedom" is $3$. If you require a circle to pass through $A,B$ then you impose two constraints, so "there is only $1$ degree of freedom left". Since our $t$ was arbitrary, that's exactly the degree of freedom left.
A: To begin with, I think that you copied the equation wrong with an extra $ = 0 $; it should be $$ x ^ 2 + y ^ 2 - 1 2 x - 4 y - 1 0 + 2 t ( 3 x + y - 1 0 ) = 0 \text . $$  And then the next line should be $$ x ^ 2 + 2 ( 3 t - 6 ) x + 2 ( - 2 + t ) y - ( 1 0 + 2 0 t ) = 0 $$ with $ 2 0 t $ instead of $ 2 t $.
The $ t $ here is called a Lagrange multiplier (after Joseph-Louis Lagrange, although he was not the first to use them); it's often written $ \lambda $ (which I think stands for Lagrange, although I'm not sure).  This is often taught in a multivariable Calculus course to find extreme values of functions of several variables under constraints; but as you can see, it has other uses.  I'll continue to use $ t $ in this answer.
As you can see from the question linked in the comment by @RamanujanXV (where it is called $ \lambda $, and the $ \lambda $ there is actually half of your $ t $), what this does is give us the equation for a whole family of circles, one circle for each value of $ t $.  When $ t = 0 $, we get the original circle $ S $; but for other values of $ t $ we get other circles; and as $ t \to \infty $, the circle gets closer and closer to the line $ L $.  (There are versions of this technique where you put a multiplier in front of each expression, so $ s ( x ^ 2 + y ^ 2 - 1 2 x - 4 y - 1 0 ) + t ( 3 x + y - 1 0 ) = 0 $, so that you can recover both original curves exactly for certain values of the multipliers.  But it's usually enough just to have one multiplier; and in this case, it's convenient not to have $ s $ so that you can say that every one of these curves is a circle.)
Now, every one of these circles passes through the point where $ S $ intersects $ L $.  This is by design, because the equation is $$ [ \text {expression which is $ 0 $ on $ S $} ] + 2 t [ \text {expression which is $ 0 $ on $ L $} ] = 0 \text , $$ so any point that is on both $ S $ and $ L $ will give $ 0 + 2 t 0 = 0 $.  You are asked to find a circle through that point with the same radius as $ S $, and here you have a whole family of circles through that point, so you just have to figure out the radius (as a function of $ t $) and solve for $ t $.  Of course, one of the solutions gives you $ S $ again, so you pick the other solution.
Hopefully your instructor would explain all of this, but once you get used to it, you just set up $$ [ \text {first equation} ] + t [ \text {second equation} ] = 0 $$ to get a family of related curves through the points where two original curves meet, without thinking very much about it.
(Your professor also used $ 2 t $ instead of $ t $, which I think was just to make it easier to complete the square when calculating the radius.  I don't know any systematic reason to use $ 2 t $.  But since the line is given equally well by $ 3 x + y - 1 0 = 0 $ or by $ 2 ( 3 x + y - 1 ) = 0 $, it makes no difference in the end.)
