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luda_lava [24]
3 years ago
13

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?

Mathematics
1 answer:
Rashid [163]3 years ago
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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<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

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<u>Step-by-step explanation:</u>

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<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

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<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

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<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

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