In the figure given below, PA and PB are tangents to the circle from point P. Which of the following is the correct relation between PA and PB?

P A > P B

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P A < P B

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P A = P B

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P A^{2} = P B^{2} + r^{2}

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Solution

The correct option is C $P A = P B$

Join the points OP, OA and OB
Now, in triangles OPA and OPB,
OP = OP (common side)
OA = OB (radii)
$∠ O A P = ∠ O B P = 90^{\circ}$ (Tangent is perpendicular to the radius)
Hence, $Δ O A P ≅ Δ O B P$ (RHS congruency)
Hence, PA = PB (CPCT)

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In the figure given below, PA and PB are tangents to the circle from point P. Which of the following is the correct relation between PA and PB?

The correct option is C PA=PB Join the points OP, OA and OB Now, in triangles OPA and OPB, OP = OP (common side) OA = OB (radii) ∠OAP=∠OBP=90∘ (Tangent is perpendicular to the radius) Hence, ΔOAP≅ΔOBP (RHS congruency) Hence, PA = PB (CPCT)

The correct option is C $P A = P B$

Join the points OP, OA and OB
Now, in triangles OPA and OPB,
OP = OP (common side)
OA = OB (radii)
$∠ O A P = ∠ O B P = 90^{\circ}$ (Tangent is perpendicular to the radius)
Hence, $Δ O A P ≅ Δ O B P$ (RHS congruency)
Hence, PA = PB (CPCT)