If C1:x2+y2−20x+64=0 and C2:x2+y2+30x+144=0. The line PQ touches the circle touches C1 at P and C2 at Q. Then the length of the shortest line segment PQ 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 A20 The equation of the first circle is (x−10)2+y2=62 The equation of the second circle is (x+15)2+y2=92 The distance between their centres d is 10−(−15)=25 If length of traverse common tangent is t1, then d2=(r1+r2)2+t21 ⇒t1=√252−152=20 If length of direct common tangent is t2, then d2=(r2−r1)2+t22 ⇒t2=√252−32=√616 ∴t1<t2⇒PQmin.=20