- Thread starter nad081
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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.

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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?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.

Thanks in advance

Thanks.

-Dan

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Yes: https://mathworld.wolfram.com/OrthogonalCurves.htmlThis 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

Thanks a million. I thought I was doing something wrong ... thanks for clarifying it. Much appreciated.

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.

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Oh, come on. Draw a picture. Two circles are orthogonal if and only if the circles share one point.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.

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

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That's tangent, not orthogonal.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

Here are two of the infinitely many solutions:

Oh, come on. You don’t know the definition of orthogonality mate! Thanks anywayOh, 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

Yes the center of the circle lies on x=5/2. Thanks againThat's tangent, not orthogonal.

Here are two of the infinitely many solutions:

View attachment 19960