Given a locus of points, show that it is a circle (find its radius and center)

In summary, the conversation discusses the locus of points on the line y=a and a linear fractional transformation, and how they relate to circles on the rectangular and complex planes. It is shown that the locus of points on the line y=a is a circle with radius 1/(2a) and center at -i/(2a). The conversation also discusses finding the center and radius of a circle given the locus of points, and it is shown that the locus of points {1/z: z E(epsilon) C} is another circle with center at the origin and radius 1/r.
  • #1
Goomba
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0
1. Suppose L is the line y=a. Show that the locus of points {1/z: z E(epsilon) L} is the circle of radius 1/(2a) with center at -i/(2a).



2. T(z)=(az+b)/(cz+d), ad != bc. T(z) is a linear fractional transformation. Linear fractional transformations take circles to circles, where the class of circles includes regular circles and straight lines.

Equation for circle on the rectangular plane: (x-a)^2+(y-b)^2=r^2 where r is the radius and (a,b) is the center.

Equation for circle on the complex plane: |z-center|=radius, I think??




3. The preceding problem with my solution:
Let C be the circle |z|=r. Show that the locus {1/z: z E(epsilon) C} is another circle with center at the origin. Find its radius.

Let z=a+bi.
Given: |z|=r
r=sqrt(a^2+b^2)

|1/z| = |1/(a+bi)| = |(a-bi)/(a^2+b^2)| = |a/(a^2+b^2) - ib/(a^2+b^2)| = sqrt{[a/(a^2+b^2)]^2 + [-b/(a^2+b^2)]^2} = sqrt[(a^2+b^2)/(a^2+b^2)^2] = sqrt[1/(a^2+b^2)] = 1/r, which is its radius.

I don't know how to show its center as the origin. Or even if I correctly found the radius here. The whole process of finding center and radius given a locus of points confuses me...
 
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  • #2
Goomba said:
1. Suppose L is the line y=a. Show that the locus of points {1/z: z E(epsilon) L} is the circle of radius 1/(2a) with center at -i/(2a).



2. T(z)=(az+b)/(cz+d), ad != bc. T(z) is a linear fractional transformation. Linear fractional transformations take circles to circles, where the class of circles includes regular circles and straight lines.

Equation for circle on the rectangular plane: (x-a)^2+(y-b)^2=r^2 where r is the radius and (a,b) is the center.

Equation for circle on the complex plane: |z-center|=radius, I think??




3. The preceding problem with my solution:
Let C be the circle |z|=r. Show that the locus {1/z: z E(epsilon) C} is another circle with center at the origin. Find its radius.

Let z=a+bi.
Given: |z|=r
r=sqrt(a^2+b^2)

|1/z| = |1/(a+bi)| = |(a-bi)/(a^2+b^2)| = |a/(a^2+b^2) - ib/(a^2+b^2)| = sqrt{[a/(a^2+b^2)]^2 + [-b/(a^2+b^2)]^2} = sqrt[(a^2+b^2)/(a^2+b^2)^2] = sqrt[1/(a^2+b^2)] = 1/r, which is its radius.

I don't know how to show its center as the origin. Or even if I correctly found the radius here. The whole process of finding center and radius given a locus of points confuses me...
You found the radius by taking |1/z|, that is, the distance from the point 1/z to the origin. In other words, you have shown that the distance from any point to the origin is a constant. Doesn't that show that the locus is a circle with the origin as center?
 

1. How do you know if a locus of points is a circle?

One way to determine if a locus of points is a circle is to check if all the points are equidistant from a fixed point, called the center. If this is true, then the locus of points is a circle.

2. What is the formula for finding the radius of a circle from a given locus of points?

The formula for finding the radius of a circle from a given locus of points is to take any two points on the circle and find the distance between them. This distance will be the radius of the circle.

3. Can you use any two points on the circle to find its radius?

Yes, any two points on the circle can be used to find the radius. This is because all points on the circle are equidistant from the center, so the distance between any two points will be the same.

4. How do you find the center of a circle from a given locus of points?

To find the center of a circle from a given locus of points, you can take any three non-collinear points on the circle and find the intersection of the perpendicular bisectors of each line segment connecting the points. The point of intersection will be the center of the circle.

5. Are there any other methods for proving that a locus of points is a circle?

Yes, there are other methods for proving that a locus of points is a circle, such as using the equation of a circle (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius. Additionally, if the locus of points is described by a quadratic equation, it can be shown to be a circle if the coefficients of x^2 and y^2 are equal and the coefficients of x and y are both 0.

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