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I have found the derivative of the graph, which is 1/√x. I know that the equation of the circle is X^2 + Y^2=r^2 where r is the radius so the equation of each circle is X^2 + Y^2=18. That is all I can get. Please help!

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- #1

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I have found the derivative of the graph, which is 1/√x. I know that the equation of the circle is X^2 + Y^2=r^2 where r is the radius so the equation of each circle is X^2 + Y^2=18. That is all I can get. Please help!

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HallsofIvy

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Okay, what is that at (1, 2)?Two circles of radius 3√2 are tangent to the the graph y^2 =4x at (1,2). Find the equation of each circle.

I have found the derivative of the graph, which is 1/√x.

No, youI know that the equation of the circle is X^2 + Y^2=r^2 where r is the radius so the equation of each circle is X^2 + Y^2=18.

If you have found the derivativeThat is all I can get. Please help!

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I plugged in x=1 into the derivative so the slope of my tangent is 1. Then I found the equation of the tangent which is y=x + 1. I now know that the slope of my circles is -1 and I know that

(x-a)^2 + (y-b)^2= 18

Could you help me with the next step??? In other words, how do I find a point with that slope and a distance or radius of 3sqrt2?

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The slope of the circles is not -1; this is the slope of the line perpendicular to the tangent.

I plugged in x=1 into the derivative so the slope of my tangent is 1. Then I found the equation of the tangent which is y=x + 1. I now know that the slope of my circles is -1 and I know that

(x-a)^2 + (y-b)^2= 18

Could you help me with the next step??? In other words, how do I find a point with that slope and a distance or radius of 3sqrt2?

Once you've found the equation of this perpendicular line...

The cool thing is that the slope of the perpendicular line is -1 so the change in the x and change in the y are the same. Hint: If you draw a triangle you will see that it is a special triangle. From here, solve for the side(s) of that triangle.

Once you've done that, just simply add/subtract to (1,2) and you will get the centers of both circles. Finally, solve for the equation of the circles.

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HallsofIvy

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Set that equal to [itex]3\sqrt{2}[/itex] and solve for a.

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