 Dayanara Terry

2022-07-02

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. Miguidi4y

Step 1
Let A,B be the intersection points of S and L; in other words $S\left(A\right)=S\left(B\right)=L\left(A\right)=L\left(B\right)=0$ . Then for any t we also have $\left(S+tL\right)\left(A\right)=\left(S+tL\right)\left(B\right)=0$ , thet 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 Frederick Kramer

Step 1
To begin with, I think that you copied the equation wrong with an extra $=0$ ; it should be
${x}^{2}+{y}^{2}-12x-4y-10+2t\left(3x+y-10\right)=0\text{.}$
And then the next line should be
${x}^{2}+2\left(3t-6\right)x+2\left(-2+t\right)y-\left(10+20t\right)=0$
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.
When $t=0$ , we get the original circle S; but for other values of t we get other circles; and as $t\to \mathrm{\infty }$ , 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
$s\left({x}^{2}+{y}^{2}-12x-4y-10\right)+t\left(3x+y-10\right)=0$
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

so any point that is on both S and L will give
$0+2t0=0$
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
$\left[\text{first equation}\right]+t\left[\text{second equation}\right]=0$
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 2t 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 2t. But since the line is given equally well by
$3x+y-10=0$ or by $2\left(3x+y-1\right)=0$ , it makes no difference in the end.)

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