Question

consider a circle with center o and let p be a point on the circle. suppose l is a tangent line to the circle at p, that is l meets the circle only at p. show that \\(\overline{op}\\) is perpendicular to l.

Explanation:

Step1: Assume a Contradiction

Suppose \( \overline{OP} \) is not perpendicular to \( L \). Then, we can draw a line from \( O \) to \( L \), say \( \overline{OQ} \), where \( Q \) is a point on \( L \) and \( \overline{OQ} \perp L \). By the definition of the perpendicular distance from a point to a line, \( OQ \) is the shortest distance from \( O \) to \( L \).

Step2: Analyze the Lengths

Since \( P \) is on the circle and \( O \) is the center, \( OP \) is the radius, let's denote \( OP = r \) (radius of the circle). Now, if \( \overline{OP} \) is not perpendicular to \( L \), then \( Q
eq P \) (because we assumed the perpendicular is \( OQ \) not \( OP \)). By the Pythagorean theorem (or the fact that the perpendicular distance is the shortest), \( OQ < OP \) (because \( OQ \) is the perpendicular distance and \( OP \) is a slant distance from \( O \) to \( L \) in this assumed case). But \( OP = r \), so \( OQ < r \).

Step3: Contradict the Tangent Definition

If \( OQ < r \), then the point \( Q \) lies inside the circle (since the distance from the center to \( Q \) is less than the radius). But \( L \) is a tangent to the circle at \( P \), which means \( L \) should intersect the circle only at \( P \). If \( Q \) is inside the circle, then the line \( L \) (which passes through \( Q \)) would intersect the circle at more than one point (because a line passing through an interior point of a circle must intersect the circle at two points), which contradicts the definition of a tangent (a tangent intersects the circle at exactly one point). So our initial assumption that \( \overline{OP} \) is not perpendicular to \( L \) is wrong.

Step4: Conclude the Result

Therefore, \( \overline{OP} \) must be perpendicular to \( L \).

Answer:

\( \overline{OP} \) is perpendicular to \( L \) (proven by contradiction as shown above).