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avatar+993 

A circle passes through the point (0,1), and is tangent to the parabola y = x^2 at (2,4). Find the center of the circle. 

 

The answer should be in the form (a,b), the answer should be a fraction not decimal. smiley

 Feb 11, 2022
 #2
avatar+14080 
+5

Find the center of the circle.

 

Hello proyaop!

 

\(f(x)=x^2\\ \frac{dy}{dx}=2x=2\cdot 2=4\\ m_g=- \frac{1}{4}\ |\ P_g(2,4)\\ g(x)=-\frac{1}{4}(x-2)+4\\ g(x)= -\frac{1}{4}x+\frac{9}{2}\)

\(h(x)=\frac{4-1}{2-0}x+1\\ h(x)=\frac{3}{2}x+1\)

\(m_i=-\frac{2}{3}\ |\ P_i(1,2.5)\\ i(x)=-\frac{2}{3}(x-1)+\frac{5}{2}\\ i(x)= -\frac{2}{3}x+\frac{19}{6}\)

\(\color{blue}i(x)=g(x)\\ -\frac{2}{3}x+\frac{19}{6}=-\frac{1}{4}x+\frac{9}{2}\\ \frac{5}{12}x=-\frac{8}{6} \)

\(x_C=-\frac{16}{5}\)

\(y= -\frac{1}{4}x+\frac{9}{2}\\ y= -\frac{1}{4}\cdot (-\frac{16}{5})+\frac{9}{2}\\ \color{blue}y_C=\frac{53}{10}\)

 

The center of the circle is \(P_C(-\frac{16}{5},\frac{53}{10})\)

laugh  !

 Feb 12, 2022
 #3
avatar+993 
+1

Wow that was a great way to do it! Thank you so much Asinus! smiley

 

oh and hi

 Feb 12, 2022
edited by proyaop  Feb 12, 2022

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