Unit circle in a curve

Question

A circle with radius $1$,lies between curve $y=\sqrt{x}$ and $x$ axis.What is centre of the circle. I tried and got solution by transforming into polar coordinates but couldn't find a good approach in cartesian coordinates although I know the answer.Please help.

Answer 1

I am assuming that when you say between, you mean that the circle is tangent to both the $x$-axis and the curve $y=\sqrt{x}$.

• Suppose that the center of the circle is at $(x_0,y_0)$. Then, since the circle is a unit circle tangent to the $x$-axis, $y_0=1$. Therefore, the center of the circle is $(x_0,1)$.

• Now, we need to find a point on the curve $y=\sqrt{x}$. Suppose that this point is $(x_1,y_1)=(x_1,\sqrt{x_1})$. Since the circle is a unit circle, it follows that $$ \sqrt{(x_0-x_1)^2+(1-\sqrt{x_1})^2}=1, $$ or that $$ (x_0-x_1)^2+(1-\sqrt{x_1})^2=1. $$

• We haven't yet used tangency to find the point on the unit circle. If the circle is tangent to the curve, then the radius of the circle from the point of tangency is perpendicular to the tangent line. The slope of the tangent line is given by the derivative, i.e., the derivative of the curve is $y'=\frac{1}{2\sqrt{x}}$. Therefore, the slope at $x=x_1$ is $$\frac{1}{2\sqrt{x_1}}.$$ On the other hand the slope of the line through the intersection point and the center of the circle is $$ \frac{1-\sqrt{x_1}}{x_0-x_1}. $$ For these two to be perpendicular, they must be negative reciprocals, in other words $$ \frac{1-\sqrt{x_1}}{x_0-x_1}=-2\sqrt{x_1}. $$

Now, one can solve this system of equations. In particular, the first equation can be written as $$ (x_0-x_1)^2=1-(1-\sqrt{x_1})^2 $$ and the second can be written as $$ x_0-x_1=-\frac{1-\sqrt{x_1}}{2\sqrt{x_1}}. $$ We can substitute the second equation into the first to get $$ \left(-\frac{1-\sqrt{x_1}}{2\sqrt{x_1}}\right)^2=1-(1-\sqrt{x_1})^2. $$ Finally, using a substitution like $\sqrt{x_1}=z$ makes the notation slightly easier. Once you solve this, you can work backwards to find the actual solution.

I get that the actual center is approximately at $(4.12402367574414,1)$.

Answer 2

Just adding a little graphical interpretation. A first guess might be that the circle center is at $(4,1)$ but this is clearly not correct because the circle intersects $y=\sqrt{x}$ twice

Putting in the center Michael Burr found (approximately $(4.12402367574414,1)$ ) looks correct