Find locus of the centre of a variable circle

[cooldudeachyut]

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

My attempt: The coordinate of the points passing thought the circle will be (k₁,0) and (0,k₂) and the centre be assumed to be (h,k).

Taking the circle to be : -
(x-h)² + (y-k)² = p

Both the above points will satisfy the equation, so :-
(0-h)² + (k₂-k)² = (k₂-h)² + (0-k)²

So on solving and replacing h and k with x and y(to get a proper locus) I get a linear equation :-

x/k₂ - y/k₁ = (k₁² - k₂²)/2k₁k₂

However I'm supposed to get a hyperbola. What am I doing wrong? Please help.

 

[ksdhart]

I'm a bit confused by this because you say "...so that the intercepts are of given length k₁ and k₂" Intercepts are defined as the point where a given line/curve crosses the x-(or y)-axis, and as such they don't have a length. Or at least that's been the case in all my math classes. Perhaps your class/book teaches something different. Can you please define what the "length of an intercept" means? If possible, please post the definition straight from your book, word-for-word to eliminate possible confusion.

I have a guess as to what it might refer, so I'll work with that for now. As you know, the only way a circle can have one x-intercept is for the x-axis to be a tangent line to said circle. Similar reasoning applies to the y-intercept. Thus, it seems highly likely that if h and k were chosen at random, there would be two intercepts most of the time (we're ignoring circles with no intercepts because the problem states that the circle has intercepts). Accordingly, I believe "length of an intercept" refers to the length of the line segment connecting the two x-intercepts, and the length of the line segment connecting the two y-intercepts. I'm actually at a bit of a loss as to if that will help with anything, but if nothing else, it places a maximum value on k₁ and k₂, that being the diameter.

 

[cooldudeachyut]

My book refers to the length of a x or y intercept of a curve as the distance between origin and the point where the curve cuts the x or y axis respectively.

Also, I don't think that taking k₁ and k₂ as length between two x and y intercepts will give a hyperbola as the locus.

 

[ksdhart]

Thanks for getting back to me with the definition. Unfortunately, it doesn't seem to help any. Because the curve in question is a circle, the distance from the origin to any point on the circle is going to be the radius. That's a known property of circles, being that they're the set of all points equidistant from the origin. Thus k1 and k2 will always have the same value. So, this suggests to me that I'm either not understanding the problem, or not understanding the meaning of the term distance of an intercept. I recommend asking your instructor for clarification on terminology, to better understand what the problem is even asking.

 

[pka]

Q- A variable circle cuts x-y axes so that the intercepts are of a given length a and b. Find the locus of the centre of circle. However I'm supposed to get a hyperbola. What am I doing wrong?

I changed lengths to \(\displaystyle a~\&~b\) so the intercepts are \(\displaystyle (a,0)~\&~(0,b)\). The lengths are non-negative by definition. Locus means path here, the path of the centre.

\(\displaystyle \begin{align*}(h-a)^2+(k)^2&=r^2\\(h)^2+(k-b)^2&=r^2\\h^2-2ah+a^2+k^2&=h^2+k^2-2bk+b^2\\-2ah+a^2&=-2bk+b^2\\ \\2ah-2bk& =a^2-b^2 \end{align*}\)

If the answer "is supposed to get a hyperbola" I have no idea what whoever wrote this means.