if the xy-plane, the graph of y=x^2 and the circle with center (0,1) and radious 3 have how many points of intersection?

1 answer:

6 0

$\bf \textit{equation of a circle}\\\\
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
radius={{ r}}\\
----------\\
center\ (0,1)\to 
\begin{cases}
h=0\\
k=1
\end{cases}\\
r=3
\end{cases} 
\\\\\\
(x-0)^2+(y-1)^2=3^2\implies x^2+(y-1)^2=9
\\\\\\
(y-1)^2=9-x^2\implies y=\sqrt{9-x^2}+1\\\\
-----------------------------\\\\
\textit{where do the parabola and circle intersect? set them to equal}
\\\\\\

$

$\bf \begin{cases}
y=x^2\\\\
y=\sqrt{9-x^2}+1
\end{cases}\qquad y=y\implies x^2=\sqrt{9-x^2}+1$

solve for "x"

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Step-by-step explanation:

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Mathematically, we can show the equivalency of the two fractions as follows;

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Therefore;

$\dfrac{2}{3} = \dfrac{2 \times 2}{3 \times 2} =\dfrac{4}{6}$

4. From the shaded region of the given figure, the fractions are equivalent

Mathematically, we have;

$\dfrac{2}{3} =\dfrac{2}{3} \times 1 = \dfrac{4}{4} \times \dfrac{2}{3} = \dfrac{4 \times 2}{4 \times 3} =\dfrac{8}{12}$

Therefore;

$\dfrac{2}{3} = \dfrac{4 \times 2}{4 \times 3} =\dfrac{8}{12}$

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$\dfrac{1}{8} + \dfrac{1}{8} = \dfrac{1}{4}$

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