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Two circles of radius 4 are tangent to the graph of y^2 = 4x at the point (1, 2). Find equations of these two circles. (Enter your answers as a comma-separated list.)

Ok, so I found the slope of y^2 = 4x @ (1,2 )by dy/dx implicitly. The answer I got was 1.

2 / y | (1 , 2) | y = x + 1

How would I proceed from here? I know how to find the tangent line from a circle and a given point, but how would I do the opposite?

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The distance between the center of the circle to the touching point should be equal to 4 and perpendicular to it.

graph from desmos.com The equation of the perpendicular line will be:

\begin{equation}g(x)=-1x+b\end{equation} \begin{equation}g(1)=-1*1+b\end{equation} \begin{equation}2=-1*1+b\end{equation} \begin{equation}b=2+1\end{equation} \begin{equation}b=3\end{equation} \begin{equation}g(x)=-x+3\end{equation}

The distance between the center of the circle and the point (1,2) is exactly 4 (the radius). Using the formula of distance:

$$ (1-x_c)^2+(2-y_c)^2=4^2 $$

Where $ x_c $ and $ y_c$ are the coordinates of the center of the circle. Since the slope of the perpendicular line is -1 it means that the ratio between $\bigtriangleup x$ and $\bigtriangleup y$ is 1. Thus $1-x_c=2-y_c$. Let $a$ be $1-x_c and 2-y_c$, then $$ a^2+a^2=4^2 $$ $$ 2a^2=4^2 $$ $$ a^2=8 $$ $$ a=\pm2\surd2$$

Then $1-x_c=\pm2\surd2$ and $2-y_c=\pm2\surd2$. $$ 1-x_c=\pm2\surd2 $$ $$ x_c=1\pm2\surd2 $$ Same with $2-y_c=\pm2\surd2$. $$ y_c=2\pm2\surd2 $$

The final equations are: $$Circle1: (x-(1+2\surd2))^2+(y-(2-2\surd2))^2=4^2$$ $$Circle2: (x-(1-2\surd2))^2+(y-(2+2\surd2))^2=4^2$$

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  • $\begingroup$ Wouldn't it be (x-[1+2sqrt(2)])^2 + (y-[2-2sqrt(2)])^2 and (x-[1-2sqrt(2)])^2 + (y-[2+2sqrt(2)])^2 due to the negative slope? Otherwise, this is the correct answer. Thanks! $\endgroup$ – Sentient Nov 24 '14 at 2:37
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    $\begingroup$ Yes, sorry my bad. $\endgroup$ – RandomGuy Nov 24 '14 at 2:40
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Hint. The radius of each circle connecting its center to point $(1,2)$ is perpendicular to the tangent line that you found, so these radii have slope the negative reciprocal of $1$, that is $-1$. So the two centers belong to the line $(y-2)=-(x-1)$, and are distance $4$ from $(1,2)$, so you could find these centers, and from there the equations of the circles.

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suppose the center is $(a,b)$. as radius is $2$ the circle must be: $$ (x-a)^2+(y-b)^2 = 4 $$ considering the slope of the tangent at (1,2) gives: $$ (1-a) = -(2-b) $$ and now also using the fact that $(1,2)$ is on the circle gives: $$ 2(1-a)^2 = 4 \\ a= 1 \pm \sqrt{2} $$ and the corresponding values of $b$ are obtained from $b=3-a$

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Or you can use geometry:

Since the tangent and normal line are perpendicular by definition, construct two right triangles from the point $(1,2)$, where the normal line is acting as the hypotenuse. Because the slope of the normal line has an absolute value of one, in essence you construct an isosceles right triangle $(45-45-90)$. Given that the hypotenuse is a length of $4$, that means each leg has a length of $2\sqrt{2}$. Just move accordingly along the normal line.

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