From a point PP, two tangents PAPA and PBPB are drawn to a circle C(O,r)C(O,r). If OP=2rOP=2r, show that APBAPB is an equilateral triangle.


Answer:


Step by Step Explanation:
  1. Let OPOP meet the circle at QQ. Join OAOA and AQAQ.
    A O Q P B
  2. We know that the radius through the point of contact is perpendicular to the tangent.  So, OAAPOAP=90(i)
  3. The circle is represented as C(O,r), this means that O is the center of the circle and r is its radius. OQ=OA=r(ii) Also, we see that OP=OQ+QP.

    Substituting the value of OP and OQ in the above equation, we have OP=OQ+QP2r=r+QPQP=2rr=r Q is the mid-point of OP. [As QP=OQ=r]
  4. As, Q is the mid-point of OP,AQ is the median from the vertex A to the hypotenuse OP of the right-angled triangle AOQ.

    We know that the median on the hypotenuse of a right- angled triangle is half of its hypotenuse.
    Thus, QA=12OP=12(2r)=r. QA=OQ=QP=rOA=OQ=QA=r[Using eq (ii)]AOQ is an equilateral triangle. AOQ=60 [Each angle of an equilateral triangle is 60]AOP=60 [As AOQ and AOP is the same angle.] (iii)
  5. We know that the sum of angles of a triangle is 180.

    For AOP, AOP+OAP+APO=18060+90+APO=180 [Using eq (iii) and eq (i)]APO=1806090=30 Also, two tangents from an external point are equally inclined to the line segment joining the center to that point.
    So, APB=2APO=2×30=60(iv)
  6. The lengths of the tangents drawn from an external point to a circle are equal.
    So, PA=PBPAB=PBA [Angles opposite to equal sides are equal.]  (v)
  7. Consider PAB PAB+PBA+APB=180[Sum of angles of a triangle.]PAB+PBA+60=180[Using eq (iv)]2PAB=120[Using eq (v)]PAB=60 Similarly, PBA=60.
  8. As all the angles of the PAB measure 60, it is an equilateral triangle.

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