Circles - does this question have missing info?

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Does this question have weird or missing information? I got the centre of the circle to have coordinate (2.5, ?) Cannot find the other one.

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Dr.Peterson

Elite Member
It looks like you have done good work, and you found that there is no was to determine y. So it looks like y can be anything!

Check that out: Either construct the figure as I did with y arbitrary and see that the resulting circle will always meet the requirements, or just try some specific values of y and see what happens.

So the problem is faulty, in the sense that "the" is inappropriate (or additional information would be needed to make it so). There are infinitely many solutions.

topsquark

Full Member
View attachment 19950

Does this question have weird or missing information? I got the centre of the circle to have coordinate (2.5, ?) Cannot find the other one.

This is just a check for me. What does it mean when you say that the circles are orthogonal? Does it mean that for the two points that the circles intersect their slopes are perpendicular?

Thanks.

-Dan

Dr.Peterson

New member
M
It looks like you have done good work, and you found that there is no was to determine y. So it looks like y can be anything!

Check that out: Either construct the figure as I did with y arbitrary and see that the resulting circle will always meet the requirements, or just try some specific values of y and see what happens.

So the problem is faulty, in the sense that "the" is inappropriate (or additional information would be needed to make it so). There are infinitely many solutions.
Thanks a million. I thought I was doing something wrong ... thanks for clarifying it. Much appreciated.

pka

Elite Member
View attachment 19950
Does this question have weird or missing information? I got the centre of the circle to have coordinate (2.5, ?) Cannot find the other one.
Oh, come on. Draw a picture. Two circles are orthogonal if and only if the circles share one point.
The given circle has radius $$2$$ and center $$(0,0)$$. The required circle is $$(x-3)^2+y^2=1$$ The point of orthogonality is $$(2,0)$$.
SEE HERE

Dr.Peterson

Elite Member
Oh, come on. Draw a picture. Two circles are orthogonal if and only if the circles share one point.
The given circle has radius $$2$$ and center $$(0,0)$$. The required circle is $$(x-3)^2+y^2=1$$ The point of orthogonality is $$(2,0)$$.
SEE HERE
That's tangent, not orthogonal.

Here are two of the infinitely many solutions:

The given circle has radius $$2$$ and center $$(0,0)$$. The required circle is $$(x-3)^2+y^2=1$$ The point of orthogonality is $$(2,0)$$.