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Two circles of unequal radii have four c...

Two circles of unequal radii have four common tangents. A transverse common tangent meets the direct common tangents at the points `P` & `Q`. If length of direct tangent (between the point of contacts) is 8 then length of `PQ` is

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To solve the problem, we need to find the length of the segment \( PQ \) formed by the intersection of the transverse common tangent and the direct common tangents of two circles with unequal radii. ### Step-by-Step Solution: 1. **Understanding the Geometry**: - We have two circles with unequal radii that have four common tangents: two direct tangents and two transverse tangents. - The direct common tangents touch the circles at points \( A \) and \( B \), and the transverse common tangents meet the direct common tangents at points \( P \) and \( Q \). 2. **Given Information**: - The length of the direct tangent \( AB \) (between the points of contact) is given as \( 8 \). 3. **Setting Up the Problem**: - Let \( PA = a \) and \( PB = b \) where \( P \) is the point on the direct tangent line. - Since \( AB = 8 \), we can express the lengths in terms of \( a \) and \( b \): \[ PA + PB = 8 \implies a + b = 8 \] 4. **Finding Lengths**: - The length of segment \( PQ \) can be expressed as: \[ PQ = PA + PB = a + b \] - Since \( a + b = 8 \), we substitute: \[ PQ = 8 \] 5. **Conclusion**: - Therefore, the length of \( PQ \) is \( 8 \). ### Final Answer: The length of \( PQ \) is \( 8 \).
To solve the problem, we need to find the length of the segment \( PQ \) formed by the intersection of the transverse common tangent and the direct common tangents of two circles with unequal radii. ### Step-by-Step Solution: 1. **Understanding the Geometry**: - We have two circles with unequal radii that have four common tangents: two direct tangents and two transverse tangents. - The direct common tangents touch the circles at points \( A \) and \( B \), and the transverse common tangents meet the direct common tangents at points \( P \) and \( Q \). ...
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