## Circles and Parabola: find smallest circle touching 2 parabs

Geometric formulae, word problems, theorems and proofs, etc.
sparkle2009
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### Circles and Parabola: find smallest circle touching 2 parabs

Consider two parabolas
$y-3=(x+1)^2$ & $x-3=(y+1)^2$

Find the equation of circle of minimum radius touching both the parabolas.

stapel_eliz
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sparkle2009 wrote:Consider two parabolas
$y-3=(x+1)^2$ & $x-3=(y+1)^2$

Find the equation of circle of minimum radius touching both the parabolas.

Have you done any calculus yet? What have you tried so far?

sparkle2009
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### Re: Circles and Parabola: find smallest circle touching 2 parabs

Yes i did calculus,I don't know how to approach the sum.

stapel_eliz
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I'm not sure, but this is my thinking so far...

If you imagine removing the circle from the middle and sliding the parabolas toward each other, their symmetry tells you that they will touch (be tangent to each other) along the line y = x. This means that their slopes will have to be m = 1 at the point of tangency.

Since the circle will be tangent to each, I think we can proceed by assuming the circle to be meeting each curve where each derivative has a value of m = 1. This means that the slopes of the radii lines will be -1, so they'll lie on the same diameter, and thus will be the midpoint of the segment between the points of tangency.

Try taking the derivative of each curve (using the upper "half" of the second curve, after solving for "y="), and setting the results equal to m = 1. Solve each for the x-coordinate of the tangency point. Then plug this x-value back into the original curve to get the corresponding y-coordinate.

Then find the midpoint of the two points of tangency and the distance between the points of tangency. Divide the distance by "2" to get the radius, and then find the circle's equation.

...I think....

Martingale
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### Re:

stapel_eliz wrote:I'm not sure, but this is my thinking so far...

If you imagine removing the circle from the middle and sliding the parabolas toward each other, their symmetry tells you that they will touch (be tangent to each other) along the line y = x. This means that their slopes will have to be m = 1 at the point of tangency.

Since the circle will be tangent to each, I think we can proceed by assuming the circle to be meeting each curve where each derivative has a value of m = 1. This means that the slopes of the radii lines will be -1, so they'll lie on the same diameter, and thus will be the midpoint of the segment between the points of tangency.

Try taking the derivative of each curve (using the upper "half" of the second curve, after solving for "y="), and setting the results equal to m = 1. Solve each for the x-coordinate of the tangency point. Then plug this x-value back into the original curve to get the corresponding y-coordinate.

Then find the midpoint of the two points of tangency and the distance between the points of tangency. Divide the distance by "2" to get the radius, and then find the circle's equation.

...I think....

That's what I did