In the given figure, two circles touch each other at a point ^@C^@. Prove that the common tangent to the circles at ^@C^@ bisects the common tangent at the points ^@P^@ and ^@Q^@.
C A B P Q R


Answer:


Step by Step Explanation:
  1. We see that ^@PR^@ and ^@CR^@ are the tangents drawn from an external point ^@R^@ on the circle with center ^@A^@.
    Thus, ^@ PR = CR \space \space \space \ldots \text{(i)}^@

    Also, ^@QR^@ and ^@CR^@ are the tangents drawn from an external point ^@R^@ on the circle with center ^@B^@.
    Thus, ^@ QR = CR \space \space \space \ldots \text{(ii)} ^@
  2. From ^@ eq \space \text{(i)} ^@ and ^@ eq \space \text{(ii)} ^@, we get
    ^@ PR = QR \space \space \text{ [Both are equal to CR]} ^@

    Therefore, ^@R^@ is the midpoint of ^@PQ^@.
  3. Thus, we can say that the common tangent to the circles at ^@C^@ bisects the common tangent at the points ^@P^@ and ^@Q^@.

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