How do you construct tangents from an external point to a circle?

Steps: (1) Draw circle with centre O. (2) Mark external point P and join OP. (3) Find midpoint M of OP. (4) Draw a circle with centre M, radius = MO. (5) This circle intersects the original at T₁ and T₂. (6) PT₁ and PT₂ are the required tangents. Justification: ∠OT₁P = 90° (angle in semicircle with diameter OP), confirming OT₁ ⊥ PT₁, so PT₁ is a tangent.

Why are the two tangents from an external point equal in length?

In triangles OPT₁ and OPT₂: OT₁ = OT₂ (radii), OP = OP (common), ∠OT₁P = ∠OT₂P = 90° (tangent ⊥ radius). By RHS congruence: △OPT₁ ≅ △OPT₂. Therefore PT₁ = PT₂ by CPCT — both tangents are equal in length.

How do you divide a line segment in a given ratio using constructions?

To divide AB in ratio m:n: (1) Draw ray AX at an acute angle. (2) Mark m+n equal arcs on AX (A₁, A₂... A_{m+n}). (3) Join A_{m+n} to B. (4) Through point A_m, draw a line parallel to A_{m+n}B. (5) Where it meets AB is point C, dividing AB in ratio m:n. This uses the Basic Proportionality Theorem (Thales theorem).

Construction of Tangents to a Circle

Constructions • Class 10 Mathematics • NCERT • CBSE

To draw tangents from an external point P to a circle with centre O and radius r: (1) Join OP. (2) Find midpoint M of OP. (3) Draw a circle with centre M and radius OM. (4) This circle intersects the original circle at points T₁ and T₂. (5) PT₁ and PT₂ are the required tangents.

Key Formulas

Length of tangent: PT = √(d² − r²) where d = distance from external point, r = radius
Tangent ⊥ radius at point of contact: OT ⊥ PT
Two tangents from same point are equal: PT₁ = PT₂
∠OAP + ∠APB + ∠OBP + ∠AOB = 360° (quadrilateral OAPB)

Frequently Asked Questions

How do you construct tangents from an external point to a circle?: Steps: (1) Draw circle with centre O. (2) Mark external point P and join OP. (3) Find midpoint M of OP. (4) Draw a circle with centre M, radius = MO. (5) This circle intersects the original at T₁ and T₂. (6) PT₁ and PT₂ are the required tangents. Justification: ∠OT₁P = 90° (angle in semicircle with diameter OP), confirming OT₁ ⊥ PT₁, so PT₁ is a tangent.
Why are the two tangents from an external point equal in length?: In triangles OPT₁ and OPT₂: OT₁ = OT₂ (radii), OP = OP (common), ∠OT₁P = ∠OT₂P = 90° (tangent ⊥ radius). By RHS congruence: △OPT₁ ≅ △OPT₂. Therefore PT₁ = PT₂ by CPCT — both tangents are equal in length.
How do you divide a line segment in a given ratio using constructions?: To divide AB in ratio m:n: (1) Draw ray AX at an acute angle. (2) Mark m+n equal arcs on AX (A₁, A₂... A_{m+n}). (3) Join A_{m+n} to B. (4) Through point A_m, draw a line parallel to A_{m+n}B. (5) Where it meets AB is point C, dividing AB in ratio m:n. This uses the Basic Proportionality Theorem (Thales theorem).

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