The center of a circle is on the line $2x+3y=4$. The circle is tangent to the line $x-3y=8$, at point $(11,1)$. Find the equation of the circle in general form.
2026-03-26 11:05:18.1774523118
Analytic geometry, circles
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The perpendicular line to $x-3y=8 \iff y=\frac13x-\frac83$ is $y_p=-3x+b$, which must pass through the point $(11,1)$, hence: $y_p=-3x+34$.
The intersection of the lines is the center of the circle: $$\begin{cases} y_p=-3x+34 \\ y=-\frac23x+\frac43\end{cases} \Rightarrow (x,y)=(14,-8).$$ Hence: $$(x-14)^2+(y+8)^2=(14-11)^2+(-8-1)^2=90.$$