The common chord of two intersecting circles $\displaystyle c_{1}: &: c_{2}$ can be seen from their centres at the angles of $\displaystyle 90^{\circ}: and: 60^{\circ}$ respectively. If the distance between their centres is equal to $\displaystyle \sqrt{3}+1$, then the radii of $\displaystyle c_{1}: &: c_{2}$ are

Let the common chord have length 2a. Using the perpendicular bisector property, the line from the centre of c1 to the chord bisects the 90 degree angle, forming a 45-45-90 triangle where the distance from the centre is a and the radius R1 is a times the square root of 2. For c2, the line from its centre bisects the 60 degree angle, forming a 30-60-90 triangle where the distance from the centre is a times the square root of 3 and the radius R2 is 2a. The sum of these two distances equals the distance between the centres, so a times the square root of 2 plus a times the square root of 3 equals 1 plus the square root of 3, meaning a equals the square root of 2. Substituting a back gives R1 as the square root of 2 and R2 as 2.

• Thanks
• Wrong explanation