Help me find the point of intersection of a line and a point.

Question

Find the equation of the circle with its center at $(-1,-3)$ and tangent to the line through the point $(-2,4)$ and $(2,1)$.

The line is $3x+4y=10$ and the point is $(-1,-3)$. What's the better formula in getting the point of intersection and the distance? Thank you! ;)

My answer is $(x+1)^2 + (y+3)^2 = 25$

Answer 1

Your answer is absolutely correct. To answer your question, you already know the equation is $(x +1)^{2} + (y + 3)^{2} = r^2$, where r is the radius. This radius is nothing but the distance of the center $(-1, -3)$ from the line $3x + 4y = 10$. You probably know the formula for distance of a point from a line. Rory Daulton has already provided you with the formulae. So that's it. Hope this helps

Answer 2

For a vector approach to this problem let $u=B-A, v=C-A$. Then the vector projection of $v$ onto $u$ gives a vector from $A$ to the point of tangency $P$ with $P=A+\frac{u.v}{v.v} v$