[ Pythagorean Theorem (direct and inverse). Auxiliary similar triangles. Product of the lengths of chord segments and the lengths of secant segments

[Theorem on the lengths of the tangent and secant; product of the entire secant by its external part

Given a right triangle $A B C$ with legs $A C=3$ and $B C=4$. Through point $C$ a line is drawn outside the triangle, forming angles of $45^{\circ}$ with the legs. Find the radius of the circle passing through points $A, B$ and tangent to this line.

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Note that $AB=5$, and the height $CH=12/5$. Let $M$ be the intersection point of line $AB$ with the specified tangent, $P$ the point of tangency, and $MQ$ the perpendicular dropped from $M$ to $BC$. From the similarity of triangles $MBQ$ and $ABC$, it follows that $BQ:MQ=4:3$, and since $BQ=MQ+4$, then $BQ=16, MQ=12, BM=20$.

Let $O$ be the center of the circle, $R$ its radius, and $F$ the intersection point of $OP$ and $AB$. By the tangent-secant theorem, $MP^2=MB \cdot MA=300$.

Consider the case when $P$ and $C$ are on the same side of point $M$. From the similarity of triangles $MHC$ and $MPF$, it follows that $\frac{MF}{PF}=\frac{MC}{CH}=5\sqrt{2}$.

Let $PF=x$, then $MF=5\sqrt{2}x, 300=MP^2=MF^2+PF^2=49x^2$.

By the intersecting chords theorem, $x(2R-x)=FB \cdot FA=(MB-MF)(MF-MA)$, or

$2xR-x^2=(5\sqrt{2}x-15)(20-5\sqrt{2}x)=175\sqrt{2}x-50x^2-300$, from which

$R=\frac{175\sqrt{2}x-49x^2-300}{2x}=\frac{175\sqrt{2}x-98x^2}{2x}=\frac{7}{2}(25\sqrt{2}-14x)=\frac{7}{2}(25\sqrt{2}-20\sqrt{5})$.

The case when $P$ and $C$ are on opposite sides of $M$ is handled similarly.

## Answer

$\frac{35(5\sqrt{2} \pm 4\sqrt{3})}{2}$