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?
Best Answer
The distance between the center of the circle to the touching point should be equal to 4 and perpendicular to it.
\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: $$Circle1: (x-(1+2\surd2))^2+(y-(2-2\surd2))^2=4^2$$ $$Circle2: (x-(1-2\surd2))^2+(y-(2+2\surd2))^2=4^2$$