From a point outside a circle, the two tangents drawn are:

### Theorem: Tangents from an External Point **Given:** A circle with center $O$ and a point $P$ lying outside the circle. Two tangents $PA$ and $PB$ are drawn from $P$ to the circle, touching it at points $A$ and $B$ respectively. **To Prove:** $PA = PB$ **Construction:** Join $OA$, $OB$, and $OP$. **Step-by-Step Proof:** 1. **Identify Right Angles:** We know that the radius is perpendicular to the tangent at the point of contact. Therefore: $$\angle OAP = \angle OBP = 90^\circ$$ 2. **Compare Triangles:** In $\triangle OAP$ and $\triangle OBP$: - $OA = OB$ (Radii of the same circle) - $OP = OP$ (Common hypotenuse) - $\angle OAP = \angle OBP = 90^\circ$ 3. **Apply Congruence:** By the RHS (Right-Angle Hypotenuse Side) congruence criterion: $$\triangle OAP \cong \triangle OBP$$ 4. **Conclusion:** Since the triangles are congruent, their corresponding parts are equal (CPCT): $$PA = PB$$ Therefore, the lengths of the tangents drawn from an external point to a circle are **equal in length**.