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Question

If c1:x2+y2−20x+64=0 and c2: x2+y2+30x+144=0. Then the length of the shortest line segment PQ which is tangent to c1 at P and c2 at Q is


A
20
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B
15
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C
22
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D
27
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Solution

The correct option is A 20
x2+y220x+64=0(x10)2+y2=62
x2+y2+30x+144=0(x+15)2+y2=92
The circles are actually as shown in the figure on the cartesian plane. The transverse common tangents are the shortest tangents and hence, we need to find their length.
Now, make constructions as shown in the figure such that PQ||AR and BR||AP. Also, AB=25
in ABR, 252=152+AR2AR=20.
PQ=AR=20
This is the required solution.

849738_129422_ans_77aace5d387640d0ba665dcd06fe70e8.png

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