Find equation of circle passing through intersection of circle S = x 2 </ms

Dayanara Terry 2022-07-02 Answered
Find equation of circle passing through intersection of circle
S = x 2 + y 2 12 x 4 y 10 = 0
and
L : 3 x + y = 10 and having radius equal to that of circle S.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Miguidi4y
Answered 2022-07-03 Author has 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 + t L ) ( A ) = ( S + t L ) ( B ) = 0 , thet is S + t L 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 + a x + b y + 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

We have step-by-step solutions for your answer!

Frederick Kramer
Answered 2022-07-04 Author has 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 1 2 x 4 y 1 0 + 2 t ( 3 x + y 1 0 ) = 0 .
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 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 λ (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 , 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
[ expression which is  0  on  S ] + 2 t [ expression which is  0  on  L ] = 0 ,
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
[ first equation ] + t [ 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
3 x + y 1 0 = 0 or by 2 ( 3 x + y 1 ) = 0 , it makes no difference in the end.)

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-12-17
The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is 109.5, the characteristic angle for tetrahedral molecules.
asked 2020-11-03
Given:
θ=240
Find the value of θ in radians (in terms of π)
asked 2020-11-02
How many lines of symmetry does the figure shown at the right have?
A) 0
B) 1
C) 5
D) 10
asked 2022-06-27
Find the equation of an ellipse if its center is S(2, 1) and the edges of a triangle PQR are tangent lines to this ellipse. P(0, 0), Q(5, 0), R(0, 4).
My attempt: Let take a point on the line PQ. For example (m,0). Then we have an equation of a tangent line for this point: ( a 11 m + a 1 ) x + ( a 12 m + a 2 ) y + ( a 1 m + a ) = 0, where a 11 etc are coefficients of our ellipse: a 11 x 2 + 2 a 12 x y + a 22 y 2 + 2 a 1 x + 2 a 2 y + a = 0. Now if PQ: y = 0, then ( a 11 m + a 1 ) = 0 , a 12 m + a 2 = 1, a 1 m + a = 0 .I've tried this method for other 2 lines PR and RQ and I got 11 equations (including equations of a center)! Is there a better solution to this problem?
asked 2022-06-10
Avery and Sasha were comparing parabola graphs on their calculators. Avery had drawn y = 0.001 x 2 in the window 1000 x 1000 and 0 y 1000 , and Sasha had drawn y = x 2 in the window k x k and 0 y k . Except for scale markings on the axes, the graphs looked the same! What was the value of k?
asked 2021-08-16
Find the area enclosed by the curve x=t22t, y=t and the y-axis
asked 2022-05-26
A cylindrical hole of radius xis bored through a shere of radius R in such a way that the axis of the hole passes throiugh the centerof the sphere. Find the value of x that maximizes the complete surface area of the remaining solid. Hint: The area of a segment of height h on a shere of radius R is 2 π R h

New questions