Understanding Tangents from an External Point

The problem describes a scenario involving a circle, a point outside the circle, and two tangents drawn from that external point to the circle. A crucial property in geometry states that the lengths of tangents drawn from an external point to a circle are always equal.

In this specific problem, we are given:

• Point A is outside the circle.
• AP and AQ are tangents from point A to the circle.
• The length of tangent AP is given by the expression \(a^2 + 14\).
• The length of tangent AQ is given as \(239\).

We need to find the value of \(a\).

Applying the Tangent Length Property

According to the property of tangents from an external point, the lengths of AP and AQ must be equal.

Therefore, we can set up an equation:

$$AP = AQ$$

Substituting the given expressions for AP and AQ:

$$a^2 + 14 = 239$$

Solving for 'a'

Now, we need to solve this equation to find the value of \(a\). This is a simple algebraic equation.

First, isolate the \(a^2\) term by subtracting 14 from both sides of the equation:

$$a^2 + 14 - 14 = 239 - 14$$

$$a^2 = 225$$

Next, take the square root of both sides to find \(a\). Since lengths are typically non-negative, we will consider the positive square root:

$$a = \sqrt{225}$$

The square root of 225 is 15.

$$a = 15$$

So, the value of \(a\) is 15.

Verification

Let's quickly check if this value of \(a\) makes the tangent lengths equal:

Length of AP = \(a^2 + 14 = (15)^2 + 14 = 225 + 14 = 239\)

Length of AQ = \(239\)

Since AP = AQ = 239, our calculated value of \(a = 15\) is correct.

The value of \(a\) is 15.

Revision Table: Key Concepts Revisited

Additional Information on Circle Tangents

Tangents have several interesting properties related to circles:

• The radius drawn to the point of contact of a tangent is perpendicular to the tangent at that point. If P is the point of contact for tangent AP, then radius OP is perpendicular to AP (\(OP \perp AP\)).
• The centre of the circle O, the external point A, and the points of contact P and Q have specific relationships. The line segment OA bisects the angle ∠PAQ. Also, triangle OAP and triangle OAQ are congruent right-angled triangles (by RHS congruence: OP=OQ (radii), OA=OA (common), ∠OPA = ∠OQA = 90°).
• The distance OA can be related to the radius and tangent length using the Pythagorean theorem in the right triangle OAP or OAQ: \(OA^2 = OP^2 + AP^2\).

These properties are fundamental when dealing with problems involving tangents from an external point to a circle.