Question

Find the equation of the circle which passes through the point $(-2,1)$ and is tangent to the line $3x-2y-6=0$ at the point $(4,3)$.

Answer 1

$3x-2y-6=0$

$3-2\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{3}{2}$

$-\frac{x_2-x_0}{y_2-y_0}=\frac{3}{2}$

$x_0=\frac{3x_1^2-3x_2^2+3(y_1-y_2{)}^2+4x_2(y_2-y_1)}{6(x_1-x_2)-4y_1+4y_2}=-\frac{2}{7}$

$y_0=\frac{(x_1-x_2{)}^2+y_1^2+3(-x_1+x_2)y_2-y_2^2}{3(x_2-x_1)+2(y_1-y_2)}=\frac{41}{7}$

$r^2=(x_1-x_2{)}^2+(y_1-y_0{)}^2$

$\Rightarrow r=\frac{10\sqrt{13}}{7}$

${(x+\frac{2}{7})}^2+{(y-\frac{41}{7})}^2=\frac{1300}{49}$