# Points M and N lie on the AC side of triangle ABC at distances of 9 and 11 respectively from vertex A. Find the

**Points M and N lie on the AC side of triangle ABC at distances of 9 and 11 respectively from vertex A. Find the radius of the circle passing through points M and N and touching the ray AB if cos∠BAC = √11 / 6**

In addition, we designate key points and draw segments, as shown in the figure.

By the tangent and secant theorem, we find AD.

AD2 = AM * AN = 9 * 11 = 99

AD = √99

Consider the ADM triangle.

By the cosine theorem, we find DM:

DM2 = AD2 + AM2-2 * AD * AM * cos∠BAC = (√99) ^ 2 + 9 ^ 2-2 * √99 * 9 * √11 / 6 = 99 + 81-18 * 3√11 * √ 11/6 = 180-3 * 3 * 11 = 180-99 = 81

DM = 9

Since DM = AM = 9, it means the triangle ADM is isosceles.

Therefore, by the property of an isosceles triangle ∠BAM = ∠ADM

According to the fourth property of the angles associated with the circle, ∠ADM is equal to half the degree measure of the arc DM.

∠DOM is the central angle, therefore it is equal to the degree measure of the arc DM, i.e. twice as many as ∠ADM.

Consider the DOM triangle.

Since OD = OM = R, this triangle is isosceles.

Draw the height of the OE, as shown in the figure.

By the property of an isosceles triangle: the height of the OE is also a bisector and a median.

Therefore, ∠DOE = ∠DOM / 2 = ∠ADM = ∠BAC

We get that cos∠DOE = cos∠BAC = √11 / 6

sin∠DOE = √1-cos2∠DOE = √1- (√11 / 6) ^ 2 = √1-11 / 36 = √25 / 36 = 5/6 (basic trigonometric formula)

DE = DM / 2 = 9/2 = 4.5 (because OE is the median).

sin∠DOE = DE / DO (by definition).

5/6 = 4,5 / DO

DO = 4.5 / (5/6) = 4.5 * 6/5 = 5.4 = R

Answer: R = 5.4