Intersecting Secant and Tangent Theorem Questions with Solutions

In the figure below, \(OC\) is tangent to the circle. Find \(x = AB\), given the lengths of segments \(OC = 42\) and \(OA = 21\).

Solution:

• Apply the intersecting secant tangent theorem above to the secant \(OB\) and tangent \(OC\) to write: \[ OC^2 = OA \times OB \]
• Substitute the known and given quantities: \[ 42^2 = 21 \times (21 + x) \]
• Expand and simplify: \[ 1323 = 21x \]
• Solve for \(x\): \[ x = 63 \]

In the figure below, \(OC\) is tangent to the circle. Find \(a\) given the lengths of segments \(OC = a - 1\), \(OA = a - 4\), \(AB = a\).

Solution:

• \(OB\) is a secant and \(OC\) is tangent, hence the intersecting secant tangent theorem gives: \[ OC^2 = OA \times OB \]
• Substitute the given quantities: \[ (a - 1)^2 = (a - 4)(a - 4 + a) \]
• Expand and group like terms: \[ a^2 - 2a + 1 = 2a^2 - 12a + 16 \]
• Rewrite the above quadratic equation in standard form as follows: \[ a^2 - 10a + 15 = 0 \]
• Solve for \(a\) to obtain two solutions: \(a = 5 + \sqrt{10}\) and \(a = 5 - \sqrt{10}\).
• Note: The length of a segment must be positive. Hence \(a = 5 - \sqrt{10}\) cannot be accepted as a solution because \(OA = a - 4 = 5 - \sqrt{10} - 4 = 1 - \sqrt{10}\) is negative.
• \(a = 5 + \sqrt{10}\) is the only solution to the given question.

In the figure below, \(OC\) is tangent to the circle and \(OE\) intersects the circle at point \(D\) where \(E\) is the center of the circle. The lengths of the segments \(OC\) and \(OD\) are given by \(OD = 6\) and \(OC = r + 3\) where \(r\) is the length of the radius of the circle.
1) Find \(r\)
2) Find the lengths of the segments \(OA\) and \(AB\) such that \(AB = 2OA\)

Solution:

1) Since \(OC\) is tangent to the circle and \(E\) is the center, then \(EC\) is a radius and is perpendicular to \(OC\).

• Use the Pythagorean theorem to write: \[ (OD + DE)^2 = EC^2 + OC^2 \]
• Note that \(DE\) is a radius and substitute the given quantities: \[ (6 + r)^2 = r^2 + (r + 3)^2 \]
• Expand and write the above equation in standard form: \[ -r^2 + 6r + 27 = 0 \]
• Solve for \(r\) to obtain two solutions: \(r = -3\) and \(r = 9\), and only the solution \(r = 9\) is valid since the length of the radius is a positive quantity.

2) The use of the intersecting secant tangent theorem gives:

\[ OC^2 = OA \times OB \]

• Substitute the known and unknown quantities: \[ (9 + 3)^2 = OA \times (OA + 2OA) \]
• The above may be written as: \[ (15)^2 = 3OA^2 \]
• Solve for \(OA\): \[ OA = \dfrac{15}{\sqrt{3}}, \quad AB = 2 \times \dfrac{15}{\sqrt{3}} \]