Tangents to a Circle with Questions and Solutions

Problem Statement: In the figure below, \(MA\) and \(MB\) are tangent to the circle with center \(O\). \(MO\) cuts the circle at point \(N\) such that the length of \(MN\) is equal to \(8\) units and the length of \(MA\) is equal to \(16\).

1. Find the radius of the circle.
2. Find the size of angle \(\angle AOB\).
3. Find the area of the shaded (in blue) sector.

Solution:

1. The tangent \(MA\) makes an angle of \(90^\circ\) with the radius \(OA\), and therefore \(OAM\) is a right triangle. Using the Pythagorean theorem, we write the equation: \[ OA^2 + AM^2 = (8 + ON)^2 \] Let \(r = OA = ON\) be the radius of the circle and rewrite the above equation as: \[ r^2 + 16^2 = (8 + r)^2 \] Expand the right side of the equation: \[ r^2 + 16^2 = 8^2 + 16r + r^2 \] Group like terms, simplify, and rewrite: \[ 16^2 - 8^2 = 16r \implies 256 - 64 = 16r \implies 192 = 16r \implies r = 12 \]
2. Let us first find the size of angle \(\angle AOM\) using the tangent formula: \[ \tan \angle AOM = \frac{AM}{OA} = \frac{16}{12} = \frac{4}{3} \] Hence: \[ \angle AOM = \arctan(4/3) \] \(OM\) bisects \(\angle AOB\), hence: \[ \angle AOB = 2 \angle AOM = 2 \arctan(4/3) \]
3. Use the formula for the area of a sector to calculate the area \(A_s\) of the shaded sector: \[ A_s = \frac{1}{2} \cdot \angle AOM \cdot r^2 = \frac{1}{2} \cdot \arctan(4/3) \cdot 12^2 \approx 133.53 \, \text{square units} \]

Problem Statement: In the figure below, \(MA\) and \(MB\) are tangent to the circle with center \(O\). \(MO\) cuts the circle at point \(N\) and the length of the arc \(ANB\) is equal to \(22\) units. The radius of the circle is equal to \(8\).

Find the distance from \(N\) to \(M\).

Solution:

1. Let \(r\) be the radius of the circle. The length of the arc \(ANB\) is given by the formula: \[ S = \angle AOB \cdot r \quad (\text{where angle is in radians}) \] Solve for \(\angle AOB\): \[ \angle AOB = \frac{S}{r} = \frac{22}{8} \]
2. \(OM\) bisects angle \(\angle AOB\); hence: \[ \angle AOM = \frac{\angle AOB}{2} = \frac{22}{16} \]
3. Use the right triangle \(OAM\) to write: \[ \cos \angle AOM = \frac{r}{OM} \] Hence: \[ OM = \frac{r}{\cos \angle AOM} = \frac{8}{\cos\left(\frac{22}{16}\right)} \approx 41.12 \]
4. \[ NM = OM - r = 41.12 - 8 \approx 33.12 \, \text{units} \]

Problem Statement: In the figure below, \(MA\) and \(MB\) are tangent to the circle with center \(O\). The radius of the circle is equal to \(20\), and \(\angle AMB = 40^\circ\).

1. Find the size of angle \(\angle AOB\).
2. Find the size of angle \(\angle AOM\).
3. Find the length of segment \(AM\).
4. Find the length of segment \(OM\).
5. Find the size of angle \(\angle ACB\) where \(C\) is a point on the circle.

Solution:

1. Since \(MA\) and \(MB\) are tangent to the circle and \(OB\) and \(OA\) are radii, \(\angle MAO\) and \(\angle MBO\) are right angles (\(90^\circ\)). The sum of all interior angles in the quadrilateral \(MAOB\) is equal to \(360^\circ\), hence: \[ \angle AOB + 90^\circ + 40^\circ + 90^\circ = 360^\circ \implies \angle AOB = 140^\circ \]
2. \(OM\) bisects angle \(\angle AOB\); hence: \[ \angle AOM = \frac{\angle AOB}{2} = 70^\circ \]
3. \(AOM\) is a right triangle, hence: \[ \tan \angle AOM = \frac{AM}{OA} \implies AM = OA \cdot \tan \angle AOM = 20 \cdot \tan 70^\circ \approx 54.95 \]
4. Using the same right triangle \(AOM\): \[ \cos \angle AOM = \frac{OA}{OM} \implies OM = \frac{OA}{\cos \angle AOM} = \frac{20}{\cos 70^\circ} \approx 58.48 \]
5. Angle \(\angle ACB\) is an inscribed angle and \(\angle AOB\) is a central angle, and both angles intercept the same arc \(AB\); hence: \[ \angle ACB = \frac{1}{2} \cdot \angle AOB = \frac{1}{2} \cdot 140^\circ = 70^\circ \]