10.257. Through point $P$ of the diameter of a given circle, a chord $A B$ is drawn, forming an angle of $60^{\circ}$ with the diameter. Calculate the radius of the circle if $A P=a$ and $B P=b$.
Solution.

Let $MN$ be the diameter of the circle with center $O$, passing through point $P$ (Fig. 10.62), $\angle APO=60^{\circ}$. Drop the perpendicular $OK$ to the chord $AB$. Then $BK=\frac{1}{2}AB=\frac{a+b}{2}; PK=BK-BP=\frac{a+b}{2}-b=\frac{a-b}{2}$. In $\triangle OKP$, $\angle OKP=90^{\circ}, \angle KOP=30^{\circ}$. Therefore, $OP=2PK=a-b$. $MP=R+a-b$, $PN=R-(a-b)$, where $R$ is the desired radius of the circle. Since

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0753.jpg?height=457&width=494&top_left_y=62&top_left_x=76)

Fig. 10.62

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0753.jpg?height=422&width=720&top_left_y=106&top_left_x=540)

Fig. 10.63

$MP \cdot PN = AP \cdot PB$, then $(R+a-b)(R-(a-b))=ab; R^2-(a-b)^2=ab$; $R=\sqrt{a^2-ab+b^2}$.

Answer: $R=\sqrt{a^2-ab+b^2}$.