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.

$S={x}^{2}+{y}^{2}-12x-4y-10=0$

and

$L:3x+y=10$ and having radius equal to that of circle S.

Miguidi4y

Beginner2022-07-03Added 13 answers

Step 1

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 $(S+tL)(A)=(S+tL)(B)=0$ , 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

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 $(S+tL)(A)=(S+tL)(B)=0$ , 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

Beginner2022-07-04Added 7 answers

Step 1

To begin with, I think that you copied the equation wrong with an extra $=0$ ; it should be

${x}^{2}+{y}^{2}-12x-4y-10+2t(3x+y-10)=0\text{.}$

And then the next line should be

${x}^{2}+2(3t-6)x+2(-2+t)y-(10+20t)=0$

with 20t instead of 2t.

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

$s({x}^{2}+{y}^{2}-12x-4y-10)+t(3x+y-10)=0$

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

$[\text{expression which is}0\text{on}S]+2t[\text{expression which is}0\text{on}L]=0\text{,}$

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

$[\text{first equation}]+t[\text{second equation}]=0$

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 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 $2(3x+y-1)=0$ , it makes no difference in the end.)

To begin with, I think that you copied the equation wrong with an extra $=0$ ; it should be

${x}^{2}+{y}^{2}-12x-4y-10+2t(3x+y-10)=0\text{.}$

And then the next line should be

${x}^{2}+2(3t-6)x+2(-2+t)y-(10+20t)=0$

with 20t instead of 2t.

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

$s({x}^{2}+{y}^{2}-12x-4y-10)+t(3x+y-10)=0$

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

$[\text{expression which is}0\text{on}S]+2t[\text{expression which is}0\text{on}L]=0\text{,}$

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

$[\text{first equation}]+t[\text{second equation}]=0$

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 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 $2(3x+y-1)=0$ , it makes no difference in the end.)

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