Find the locus of center of circle

In summary: The center of the circle will be at the intersection of the bisectors. The center of the circle will be at the intersection of the bisectors.
  • #1
utkarshakash
Gold Member
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13

Homework Statement


A variable circle cuts x and y axes so that intercepts are of given length k1 and k2. Find the locus of center of circle

Homework Equations



The Attempt at a Solution


There must be four intercepts but only two are given.
 
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  • #2
hi utkarshakash! :smile:
utkarshakash said:
There must be four intercepts but only two are given.

yes, but you only need two :wink: …

get on with it! :smile:​

hint: K1K2 is a chord of the circle
 
  • #3
There is a way of interpreting this problem so that it makes sense and you get a single circle as the answer, but I'm dubious as to whether or not that is the intended interpretation of the problem.

Other than that interpreation (and, can you see the condition that allows/forces it?) then I believe you are right that to get a single circle as the answer, you need all four intercepts (actually, three will do, and two will always give you two circles as the answer if you assume that the given intercepts are just 2 of the three or 4 that exist)
 
  • #4
tiny-tim said:
hi utkarshakash! :smile:


yes, but you only need two :wink: …

get on with it! :smile:​

hint: K1K2 is a chord of the circle

I already know that. I assume the general equation of the circle and then proceed ahead. I am thinking of drawing perpendiculars from the centre on two chords. But will this help?
 
  • #5
hi utkarshakash! :wink:
utkarshakash said:
I am thinking of drawing perpendiculars from the centre on two chords. But will this help?

you only have one chord, but yes, do draw the bisector! :smile:

(i'm assuming it mean that the x intercept is always (k1,0), and the y intercept is always (0,k2))
 
  • #6
tiny-tim said:
(i'm assuming it mean that the x intercept is always (k1,0), and the y intercept is always (0,k2))

I am not taking it that way. I was thinking k1 and k2 to be two separate chords.
 
  • #7
utkarshakash said:
I am not taking it that way. I was thinking k1 and k2 to be two separate chords.

in that case, how could there be four intercepts? :confused: …
utkarshakash said:
There must be four intercepts but only two are given.

ok, then assume the centre is at (x,y) and that the radius is r,

then find the two lengths of the intercepts with the axes, put one equal to k1, the other equal to k2, and solve for x and y …

what do you get? :smile:​
 
  • #8
The only way I can make sense of the question is to take k1 and k2 to be the distances between intercepts on the respective axes, not the coordinates of the intercepts. That way you do need all four.
utkarshakash, is that what you meant?
 
  • #9
haruspex said:
The only way I can make sense of the question is to take k1 and k2 to be the distances between intercepts on the respective axes, not the coordinates of the intercepts. That way you do need all four.
utkarshakash, is that what you meant?

Yes, exactly what I meant.:approve: But tiny-tim is taking it the other way. He is assuming k1 and k2 to be the coordinates of the intercepts on x and y axes. Which one is correct?
 
  • #10
no, I've taken it either way!

for your way …
tiny-tim said:
ok, then assume the centre is at (x,y) and that the radius is r,

then find the two lengths of the intercepts with the axes, put one equal to k1, the other equal to k2, and solve for x and y …

what do you get? :smile:​
 
  • #11
tiny-tim said:
no, I've taken it either way!

for your way …

I just made a mistake in understanding the question. Actually k1k2 is a chord of the circle as you said and not that they are 2 separate chords.
 
  • #12
utkarshakash said:
I just made a mistake in understanding the question. Actually k1k2 is a chord of the circle as you said and not that they are 2 separate chords.

ok, in that case the first thing is to draw the bisector of the chord :wink:
 
  • #13
tiny-tim said:
ok, in that case the first thing is to draw the bisector of the chord :wink:

By drawing the bisector and joining the center and (0,k2) I get a right triangle. But my idea is to calculate the distance between centre and the given two points and equate them. But that way I get the wrong answer.
 
  • #14
utkarshakash said:
… my idea is to calculate the distance between centre and the given two points and equate them.

i'm not following this :redface: …

if it's on the bisector, why do you need to check that? :confused:​
 
  • #15
tiny-tim said:
i'm not following this :redface: …

if it's on the bisector, why do you need to check that? :confused:​

Ok so what do you want me to do after drawing the bisectors?
 
  • #16
utkarshakash said:
Ok so what do you want me to do after drawing the bisectors?

(bisectors? i thought there was only one chord now? :confused:)

well, where will the centre be in relation to the bisector?
 
  • #17
tiny-tim said:
(bisectors? i thought there was only one chord now? :confused:)

well, where will the centre be in relation to the bisector?

Sorry it was a typo. I can find the equation of line passing through the mid point of chord and perpendicular to it. Next I can satisfy the centre in the equation. Am I correct?
 
  • #18
utkarshakash said:
Next I can satisfy the centre in the equation.

do you mean that the centre will lie on the line, and so the centre will satisfy the equation?

yes :smile:
 

FAQ: Find the locus of center of circle

What is the definition of "locus of center of circle"?

The locus of center of circle refers to the set of points that are equidistant from a fixed point, known as the center, on a plane.

How can you find the locus of center of circle?

To find the locus of center of circle, you can use the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) is equal to √[(x2-x1)^2 + (y2-y1)^2]. This formula can be applied to the center point and any other point on the circle to determine the locus.

What does the locus of center of circle represent?

The locus of center of circle represents the geometric shape of the circle. It is the boundary that defines all the possible points that can be the center of the circle.

Can the locus of center of circle be a line?

Yes, the locus of center of circle can be a line in special cases. When the radius of the circle is equal to 0, the locus of center of circle will be a single point. When the radius is greater than 0, the locus of center of circle will be a circle. However, in some cases, such as when the radius is equal to infinity, the locus of center of circle will be a line.

How is the locus of center of circle related to the equation of a circle?

The equation of a circle, (x-h)^2 + (y-k)^2 = r^2, represents the locus of center of circle with center (h,k) and radius r. This equation can be used to find the center and radius of a circle given its locus, or vice versa.

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