MHB Ernesto's question at Yahoo Answers regarding finding a locus of points

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Points
AI Thread Summary
The problem involves finding the locus of points P from which the sum of the lengths of tangents to two circles, C1 and C2, is constant. The circles are defined by the equations x² + y² = 4 and x² + y² = 9, with radii r1 = 2 and r2 = 3, respectively. The sum of the tangent lengths from point P is given as S = 5. By applying the tangent length formula and manipulating the equations, the locus of points is derived to be a circle with the equation x² + y² = (√145/4)². This indicates that the locus is a circle centered at the origin with a specific radius.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

The sum of the lengths of the tangents from a point P to the circumferences...?


The sum of the lengths of the tangents from a point P to the circumferences:

C1 : x² + y² = 4

and

C2 : x² + y² = 9

is constant and equal to 5.
Determine the locus of the point P.

Explain.

Thanks.

I have posted a link there to this thread so the OP can view my work.
 
Mathematics news on Phys.org
Hello Ernesto,

Let's orient a circle or radius $r$ centered at the origin, and place a point on the $x$-axis outside of the circle, and let it's distance from the origin be $d$. Drawing a tangent line from this point to the circle (in the first quadrant), we form a right triangle, since the tangent line is perpendicular to the radius of the circle drawn to the tangent point. Let $\ell$ be the length of this line segment. Hence, we may state:

$$r^2+\ell^2=d^2$$

Solving for $\ell$, we obtain:

$$\ell=\sqrt{d^2-r^2}$$

Now, observing that there is also a fourth quadrant tangent point, we may state that the sum $S$ of the lengths of the tangent line segments is:

$$S=2\sqrt{d^2-r^2}$$

Now, if we do this for two circles, one of radius $r_1$ and the other of radius $r_2$, we then have:

$$S=2\left(\sqrt{d^2-r_1^2}+\sqrt{d^2-r_2^2} \right)$$

In order to ease solving for $d$, let's write the equation as:

$$S-2\sqrt{d^2-r_2^2}=2\sqrt{d^2-r_1^2}$$

Squaring both sides of the equation, there results:

$$S^2-4S\sqrt{d^2-r_2^2}+4\left(d^2-r_2^2 \right)=4\left(d^2-r_1^2 \right)$$

Distributing and combining like terms, and isolating the term with the remaining radical we obtain:

$$4S\sqrt{d^2-r_2^2}=4r_1^2-4r_2^2+S^2$$

Squaring again, we obtain:

$$16S^2\left(d^2-r_2^2 \right)=\left(4r_1^2-4r_2^2+S^2 \right)^2$$

$$d^2=\frac{\left(4r_1^2-4r_2^2+S^2 \right)^2}{16S^2}+r_2^2=\frac{\left(4\left(r_1^2-r_2^2 \right)+S^2 \right)^2+\left(4r_2S \right)^2}{(4S)^2}$$

Now, since this will hold for any point that is a distance $d$ from the origin, which describes a circle centered at the origin having radius $d$, we then find the locus of points is:

$$x^2+y^2=\frac{\left(4\left(r_1^2-r_2^2 \right)+S^2 \right)^2+\left(4r_2S \right)^2}{(4S)^2}$$

Now, using the data given with the problem:

$$r_1=2,\,r_2=3,\,S=5$$

We find the locus of points is:

$$x^2+y^2=\left(\frac{\sqrt{145}}{4} \right)^2$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top