I'm confuse on solving this problem:
Find the equation of the circle which is tangent to the line $3x-2y=5$ at $(3,2)$ and it is passing through $(-2,1)$.
Can you pretty answer and explain how did you do the solution?
Please. I barely need to understand this lesson a lot better. Thank you.
Hint. Let $(x_0,y_0)$ be the centre of such circle. Then
i) $(x_0,y_0)$ stays on the line orthogonal to $3x-2y=5$ at $(3,2)$: $$2(x-3)+3(y-2)=0$$ that is $$2x+3y=12.$$ ii) $(x_0,y_0)$ stays on the perpendicular bisector of the segment of extreme points $(3,2)$ and $(-2,1)$: $$(x-3)^2+(y-2)^2=(x+2)^2+(y-1)^2$$
that is $$5x+y=4$$ Can you take it from here?