# Find 'k' in Circle Inscribed in a Triangle

• zorro
In summary, the circle x2 + y2 - 4x - 4y + 4 = 0 is inscribed in a triangle, which has two of its sides along the coordinate axes. If the locus of the circumcentre is of the form x + y - xy + k(x2 + y2)1/2= 0, find k.f

## Homework Statement

The circle x2 + y2 - 4x - 4y + 4 = 0 is inscribed in a triangle, which has two of its sides along the coordinate axes. If the locus of the circumcentre is of the form
x + y - xy + k(x2 + y2)1/2= 0. Find k.

## The Attempt at a Solution

The centre of the given circle is (2,2) and its radius is 2 units. Let the circumcentre be (h,k).
The intercepts of one side of the triangle on x and y axes are 2h and 2k respectively.

The distance between incentre and circumcentre is (R2 - 2Rr)1/2, where R = (h2 + k2)1/2
Square of it must be equal to (h -2)2 + (k -2)2

i.e. (R2 - 2Rr) = (h -2)2 + (k -2)2

On solving I got

h + k - (h2 + k2)1/2 = 2

which is not of the form given.
Please point out my mistake if any.

So far as I can tell, the question doesn't imply that that IS the form of the locus. It says, IF that is the form, what is k?

What are you trying to say? I don't get you. The question means that if we find out the locus of the circumcentre, it will be of the form given.

Hi Abdul! The circle x2 + y2 - 4x - 4y + 4 = 0 is inscribed in a triangle, which has two of its sides along the coordinate axes. If the locus of the circumcentre is of the form
x + y - xy + k(x2 + y2)1/2= 0. Find k.

don't you mean that it has two of its vertices on the coordinate axes? Let the circumcentre be (h,k).
The intercepts of one side of the triangle on x and y axes are 2h and 2k respectively.

how do you get that?
So far as I can tell, the question doesn't imply that that IS the form of the locus. It says, IF that is the form, what is k?

vertigo, that's not helpful don't you mean that it has two of its vertices on the coordinate axes? Yes.

how do you get that?

For a right angled triangle, the circumcentre lies on the mid-point of the hypotenuse.
Let the vertices of the triangle on x and y axes be (x,0) and (0,y) resp.
Using the midpoint formula, we get x=2h and y=2k, which are the intercepts of one side of the triangle (hypotenuse) don't you mean that it has two of its vertices on the coordinate axes? ## Homework Statement

The circle x2 + y2 - 4x - 4y + 4 = 0 is inscribed in a triangle, which has two of its sides along the coordinate axes.

I think you might have read it as the circle is inscribed in the triangle.

Abdul Quadeer, why must that distance be equal to the other distance? This assumption is saying that the circumcentre lies on the circumference of the circle. This isn't true.

Abdul Quadeer, why must that distance be equal to the other distance? This assumption is saying that the circumcentre lies on the circumference of the circle. This isn't true.

I am not equating the distance between the circumcentre and the incentre to 2 units. (h,k) is any point on the hypotenuse which may or may not lie on the circumference of the incircle.

Tim is right, I did not explain the nuance of language that I think this question revolves around in a helpful way.

So, what does the question need to be answered? Is it asking for one to find a form for the locus? Well, no, because we are given a form. So calculating a form is not part of the question. The question is saying, from this locus form, find k.

To do that, we only need one circumcentre to substitute into that locus form, to find k. The easiest circumcentre to use is the one on the line y = -x.

The nuance of language is this. Because the question says "IF this is the form, find k", that locus form may not actually be the locus form. Logically, it means that one must pretend in the current context that that is the locus form, whether or not it actually is the form. We are pretending that it is.

So I think the question was about being able to see from the circle equation where the circumcentre on the line y = -x is, and substituting those coordinates into the "pretend" locus form.

I'll just add, in my initial reply, I meant to indicate that a "mistake" that was made was a mistake of understanding the language. But Tim is correct in that I should have foreseen that it was a language nuance that needed to be explained. I don't know what the form of the locus is, I haven't worked it out. I just wanted to indicate that language mistake.

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To do that, we only need one circumcentre to substitute into that locus form, to find k. The easiest circumcentre to use is the one on the line y = -x.

The parameter 'k' is uniquely defined for any (x,y).

The easiest circumcentre to use is the one on the line y = -x.

How does the circumcentre lie on y= -x?
Refer the figure. #### Attachments

How does the circumcentre lie on y= -x?

Sorry, I meant y = x. I was thinking of a sketch I made that was reversed because I wasn't thinking in terms of coordinates at the time.

On the other point, if k is unique depending on which circumcentre you choose to substitute, I can only think that whoever wrote the question assumed that everyone would substitute in the one I suggested. I can't make sense of it any other way. I trust your calculations, Abdul.

Consider this question-

A straight line segment of length 'p' units moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line in the ratio 1:2 .

If you solve this, you get

x2/4 + y2 = p2/9

Now I will change the question.

A straight line segment of length 'p' units moves with its ends on two mutually perpendicular lines. If the equation of the locus of the point which divides the line in the ratio 1:2 is
x2/k + y2 = p2/9, find the value of 'k'.

Do you get the point there?

I don't really know what the second question is asking. It is confusing because it seems to want one to calculate the formula but it also gives one the formula. But again, I would assume that the right way to answer it is to substitite one point into that formula and solve for k, and if that formula was incorrect, deriving the correct formula would take long in the exam and would not be given marks. I am thinking in an exam context: being given the formula, one need only substitute into it.

And let's look at it the other way. Suppose that formula is wrong. Is there any way of answering the question now? It wouldn't be right to get the real formula, because where would you put k in the real formula? The right thing to do to hopefully get the marks would be either to prove that that formula is incorrect, and therefore one can't solve for k, or substitute in one point to find k. What I'm saying is, if it is correct, it can be used. If it isn't correct, it can't be fixed.

Even if whoever wrote the question intended one to derive the formula, that would have to be known from contextual knowledge, like knowing from lectures what the professor wants, or looking at how many marks have been allocated to the question. But from the language alone, there is nothing (in English, according to the use of the word "if") to suggest that one must derive the formula. Without the contextual knowledge to suggest that a derivation is required, I believe the correct way to answer such a question is to substitute in one point.

I hope this makes sense. If the locus form in the original problem is incorrect, then I can only assume either that the author of the question made a mistake, or there was a typing error.

Hi Abdul! The circle x2 + y2 - 4x - 4y + 4 = 0 is inscribed in a triangle, which has two of its sides along the coordinate axes. If the locus of the circumcentre is of the form
x + y - xy + k(x2 + y2)1/2= 0. Find k.
For a right angled triangle, the circumcentre lies on the mid-point of the hypotenuse.

Ah, yes. (Now I see your diagram, it's obvious! )

ok, now as vertigo says, since the question isn't asking you to prove the equation, it's easiest to choose just one triangle …

and the easiest one to choose will be the one with h = k, ie the one with the line parallel to x = -y, since the circumcentre of that triangle is obviously at … ? I hope this makes sense. If the locus form in the original problem is incorrect, then I can only assume either that the author of the question made a mistake, or there was a typing error.

and the easiest one to choose will be the one with h = k, ie the one with the line parallel to x = -y, since the circumcentre of that triangle is obviously at … ? Alright. I got the correct answer by substitution. I agree with vertigo that the question doesnot demand to prove it.
Now can anybody point out my mistake in finding out the equation of the required locus? or possibly suggest another approach for finding the equation?

Now can anybody point out my mistake in finding out the equation of the required locus?
Let the circumcentre be (h,k).
The intercepts of one side of the triangle on x and y axes are 2h and 2k respectively.

The distance between incentre and circumcentre is (R2 - 2Rr)1/2, where R = (h2 + k2)1/2
Square of it must be equal to (h -2)2 + (k -2)2

Your line defined by h and k is a completely general line …

you need to restrict it so that it is tangent to the circle. (in fact, that restriction will probably be an easier way of finding the locus than using the distance between the two centres)

Ah, I see. The restriction is that the perpendicular distance of the line from the centre of the circle is 2 units .

Thank you tiny-tim!
I got the answer Last edited: