Answer :

Benjamin16
[tex]\begin{cases} (x-3)^{2} + y^{2} = R^{2} \\ y = \sqrt{x} \implies y^{2} = x \end{cases} \\ \\ (x-3)^{2} + x = R^{2} \\ \\ x^{2} - 6x + 9 + x - R^{2} = 0 \\ \\ x^{2} - 5x + 9 - R^{2} = 0 \\ \\ \Delta = 0 \\ \\ \Delta = (-5)^{2} - 4 \cdot 1 \cdot (9 - R^{2}) = 25 - 36 + 4R^{2} = 4R^{2} - 11 \\ \\ 4R^{2} - 11 = 0 \\ \\ R^{2} - \dfrac{11}{4} = 0 \\ \\ R^{2} = \dfrac{11}{4} [/tex]
[tex](x-3)^{2} + x = \dfrac{11}{4} \\ \\ x^{2} - 6x + 9 + x - \dfrac{11}{4} = 0 \\ \\ x^{2} - 5x + \dfrac{25}{4} = 0 \\ \left (x - \dfrac{5}{2} \right)^{2} = 0 \\ \\ x = \dfrac{5}{2} \\ \\ y = \sqrt{ \dfrac{5}{2} } = \dfrac{\sqrt{10}}{2} \\ \\ \boxed{P = \left ( \frac{5}{2}, \dfrac{\sqrt{10}}{2} \right ) }[/tex]
meerkat18

Distance between (x,y) and (3,0) is:

D = sqrt{ (x-3)^2 + y^2 }

y = sqrt(x)

y^2 = x

Then substitute D,  that would be

D = sqrt{ (x-3)^2 + x } = sqrt(x^2 - 6x + 9 + x}

D = sqrt(x^2 - 5x + 9)

Minimize D by finding the derivative with respect to x, equating it to 0 and solving for x.

D = sqrt(x^2 - 5x + 9)

find derivative and set it to 0 in order to minimize D

D' = 1/2(x^2 - 5x + 9)^(-1/2) * (2x - 5) = 0

D' will be zero when numerator = 0

2x - 5 = 0

x = 5/2

Put x = 5/2 back in y=sqrt{x}

y = sqrt(5/2)

 

the points of the curve that is nearest to  (3,0) will be (5/2, sqrt(5/2))

So the answer will be  (2.5, 1.58)

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