Tangents to a Circle with Questions and Solutions

Tangent to a Circle

A tangent to circle touches the circle at one point only. In the figure below the tangent \( T \) cuts the circle at point \( P \) called the point of tangency.
one tangent to a circle
An important property of the tangent to a circle it that the tangent \( T \) and the radius \( OP \) are perpendicular.

Two Intersecting Tangents to a Circle

In the figure below, \( MA \) and \( MB \) are tangent to the same circle with center \( O \). The two tangents intersect at point \( M \)
two intersecting tangents to a circle
The most important properties are

  1. \( MO \) is an angle bisector to \( \angle AMB \)
  2. \( OM \) is an angle bisector to \( \angle AOB \)
  3. the right triangles \( \triangle AMO \) and \( \triangle BMO \) are congruent.



Questions With Solutions

Question 1
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 \( AOB \).
3) Find the area of the shaded (in blue) sector.
two intersecting tangents to a circle question 1
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: \( \quad OA^2 + AM^2 = (8 + ON)^2 \)
Let \( r = OA = ON \) be the radius of the circle and rewrite the above equation as: \( \quad r^2 + 16^2 = (8 + r)^2 \)
Expand the right side of the equation: \( \quad r^2 + 16^2 = 8^2 + 16 r + r^2 \)
Group like terms, simplify and rewrite the above equation as: \( \quad 16^2 - 8^2 = 16 r \)
Solve the above for \( r \) to obtain: \( \quad r = 12 \)
2) Let us first find the size of angle \( AOM \) using the tangent formula: \( \quad \tan \angle AOM = \dfrac{AM }{OA} = \dfrac{16}{12} = \dfrac{4}{3} \)
Hence: \( \quad \angle AOM = \arctan (4/3) \)
\( OM \) bisects \( \angle AOB \), hence \( \quad \angle AOB = 2 \angle AOM = 2 \arctan(4/3) \)
3) Use the formula of the area of a sector to calculate the area \( A_s \) of the shaded sector as: \( \quad A_s = (1/2) \times \angle AOM \times r^2 = (1/2) \times 2 \arctan(4/3) \times 12^2 \approx 133.53\) square units



Question 2
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 \( M \).
two intersecting tangents to a circle question 2
Solution
Let \( r \) be the radius of the circle. The length of the arc \( ANB \) is given by the formula: \( S = \angle AOB \times r\)
Solve the above for \( \angle AOB \): \( \quad \angle AOB = \dfrac{S}{r} = \dfrac{22}{8} \)
\( OM \) bisects angle \( AOB \); hence \( \quad \angle AOM = \dfrac{\angle AOB}{2} = \dfrac{22}{16} \)
Use the right triangle \( OAM \) to write: \( \cos \angle AOM = \dfrac{r}{OM} \)
Hence: \( OM = \dfrac{r}{\cos \angle AOM } = \dfrac{8}{\cos \dfrac{22}{16}} \approx 41.12\)
\( NM = OM - r = 41.12 - 8 \approx 33.12 \) units



Question 3
In the figure below, \( MA \) and \( MB \) are tangent to the circle with center \( O \). The radius of the circle is equal to \( 20 \).
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.
two intersecting tangents to a circle question 3
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.
The sum of all interior angles in the quadrilateral \( MAOB \) is equal to \( 360^{\circ} \), hence: \( \quad \angle AOB + 90^{\circ} + 40^{\circ} + 90^{\circ} = 360^{\circ} \)
Solve the above to obtain: \( \quad \angle AOB = 140^{\circ} \)
2) \( OM \) bisects angle \( \angle AOB \); hence \( \angle AOM = \dfrac {\angle AOB}{2} = 70^{\circ}\)
3) \( AOM \) is a right triangle with right, hence \( \quad \tan \angle AOM = \dfrac{AM}{OA} \)
hence, \( \quad AM = OA \times \tan \angle AOM = 20 \times \tan 70^{\circ} = 54.95 \)
4) Using the same triangle as in part 3), \( \quad \cos \angle AOM = \dfrac{OA}{OM} \)
hence, \( \quad OM = \dfrac{OA}{\cos \angle AOM} = \dfrac{20}{\cos \angle 70^{\circ} } = 58.48 \)
5) Angle \( \angle ACB \) is an inscribed angle and \( \angle AOB \) is a central angle and both angles intercept the same arc \( AB \); hence \( \angle ACB = (1/2) \times AOB = 70^{\circ}\)



More References and Links

The Four Pillars of Geometry - John Stillwell - Springer; 2005th edition (Aug. 9 2005) - ISBN-10 : 0387255303
Geometry Tutorials, Problems and Interactive Applets
{ezoic-ad-1}
{ez_footer_ads}