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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?

ACB
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Aleph
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2 Answers2

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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.

Kushi
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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.)

Toby Bartels
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  • But why does using a multiplier work? Why does the formula $S+\lambda L =0$ work? I read both your answer an the above one but still am confused – Aleph Jan 27 '22 at 05:52
  • I'd say that the why part is the equation in the middle, [expression which is 0 on ] + 2[expression which is 0 on ] = 0, or $S+\lambda L=0$ as you (and the question linked by RamanujanXV) put it. This is what guarantees that a point that's on both the circle and the line must also be on any curve given by one of these equations. There's no guarantee that one of these equations will give you what you want (that depends on what you want), but it's worth a look, and in the case of ‹circle through the points where $S$ and $L$ intersect with the same radius as $L$›, there is one. – Toby Bartels Jan 27 '22 at 16:59
  • Maybe you want to ask: why does this give you specifically the circles through the points where $S$ and $L$ meet? I don't have a better answer for that other than that the equation is still a quadratic equation whose quadratic part is $x^2+y^2$, and we already know at least two points on it, so it's still the equation for a circle. But if you wanted to get, say, a parabola through the points where $S$ and $L$ meet, then this will not work! So it's not a perfect method to get curves through those points, just a standard way that's often useful. – Toby Bartels Jan 27 '22 at 17:04