Collision Resolving - Point outside Circle

Question

In a timestep-based simulation, a collision between a point particle p(x,y) with velocity v(x,y), that started off from inside a circle (x-a)^2 + (y-b)^2 = r^, with that circle, occurred between two timesteps, such that the point particle has already left the circle when the collision gets detected.

Therefore I want to move the particle back by amount outsideDepth(x,y) such that it lies exactly on the circle.

Now the question is: How can I determine the distance l between the point p and the intersection of the velocity vector with the circle?

In code:

Vector2 circleCollision(double a, double b, double r, double x, double y){

    double s = sqrt( pow((x-a),2) + pow((y-b),2) );

    if (s>r) {
        Vector2 outsideDepth = {0,0};

        // determine depth by which point lies outside circle as vector (x,y)

        return outsideDepth;
    }
}

EDIT Attempt at Ian's solution, substitute 2 and 3 in 1 and rearange for t, then determine p and q as follows:

p = 1/( pow(v.x,2) + pow(v.y,2) ) * (-2*x*v.x + 2*v.x*a - 2*y*v.y + 2*v.y*b);
q = 1/( pow(v.x,2) + pow(v.y,2) ) * (-2*x*a -2*y*b + x*x + y*y + a*a + b*b - r*r);

root = sqrt( pow((p/2),2) - q );
t1 = -p/2 + root;
t2 = -p/2 - root;

// ???

Answer 1

Simultaneously solve(easy code)

(x-a)^2 + (y-b)^2 = r^2

and

x = p(x) - v(x)*t

y = p(y) - v(y)*t

for some t. There may be zero, one or two solutions depending on the discriminant (use conditionals on b^2 - 4ac). If two solutions (b^2 > 4ac) choose the t which minimizes size(p(x,y) - t(x, y)) (use pythag). Return.