An Arc by Construction

Let T1, T2 = two points of tangency of tangents from A and O(0,0) = center of the circle. These three points are the special cases of 'midpoints of chords' in the problem. So the desired circle passes through them.

Triangles AT1O and AT2O are right angle triangles Angle T1OA and angle T1AO are complimentary So angles T1OT2 and T1AT2 must be supplimentary, making the quadrilateral AT1OT2 cyclic. So a circle passing through T1, O, T2 must pass through A as well!

So the center of the circle must be midpoint of (0,0) and (6,8) and the radius must be half their distance = 5

A line passing through point $A$ is given by (parametric equation):

$p = A + t u$

where $u$ is a unit vector of an arbitrary direction,

The line intersects the given circle when

$p^T p = R^2$

where $R$ is the radius of the given circle ( $R = 7$ in this problem).

Plugging in the expression for $p$ , we get,

$t^2 u^T u + 2 t u^T A + A^T A = R^2$

Since $u$ is a unit vector, this becomes

$t^2 + 2 t u^T A + A^T A = R^2$

the solutions of this quadratic equation (if they exist) are $t_1$ and $t_2$ ,

which satisfy

$t_1 t_2 = A^T A - R^2$ , and

$t_1 + t_2 = -2 u^T A$

Now the corresponding points of intersection (the end points of the secant) are $p_1 and p_2$ , where

$p_1 = A + t_1 u$ $p_2 = A + t_2 u$

The midpoint is $P = 1/2 (p_1 + p_2) = A + \dfrac{1}{2} (t_1 + t_2) u = A - (u^T A) u = (I - u u^T) A$

Let's form $P^T P$ , and see what that is equal to,

$PT P = AT (I - u u^T) (I - u u^T) A = A^T (I - u u^T) A = A^T P$

Therefore,

$PT P - A^T P = 0$

Let $P_0 = \dfrac{1}{2} A$ , then

$\boxed{ \large{ (P - P_0)^T (P - P_0) = P_0^T P_0 = \dfrac{1}{4} A^T A } }$

This is an equation of a circle with center $P_0 = 1/2 A$ and radius of $\dfrac{1}{2} \sqrt{A^T A}$ .

In this problem, A = (6, 8), so the center is (3, 4) and the radius $r = \dfrac{1}{2} \sqrt{6^2 + 8^2} = 5$ .

Therefore, the required triplet is $( 3, 4, 5 )$ .