Question

PQ is a common tangent of two circles that touches the two circles at point A and B respectively . prove that extended AB bisects PQ.?​

Answer 1

Let PQ be the common tangent through P, where Q is the point that meets AB

Now, QA=QP (Tangents to the circle from the same point)

QP=QB (Tangents to the circle from the same point)

Thus, QA=QP=QB

Thus, Points A,B,P lie on the circle with centre Q and AB as the diameter.

Thus, ∠APB=90

∘

(Angle in a semi-circle)

Answer 2

Answer:

Let PQ be the common tangent through P, where Q is the point that meets AB

Now, QA=QP (Tangents to the circle from the same point)

QP=QB (Tangents to the circle from the same point)

Thus, QA=QP=QB

Thus, Points A,B,P lie on the circle with centre Q and AB as the diameter.  

Thus, ∠APB=90  

∘

 (Angle in a semi-circle)

Step-by-step explanation:

Let PQ be the common tangent through P, where Q is the point that meets AB

Now, QA=QP (Tangents to the circle from the same point)

QP=QB (Tangents to the circle from the same point)

Thus, QA=QP=QB

Thus, Points A,B,P lie on the circle with centre Q and AB as the diameter.  

Thus, ∠APB=90  

∘

 (Angle in a semi-circle)