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Transverse Common Tangent

If C1 and C2 ...

Question

If $C_{1}$ and $C_{2}$ are circles whose equations are $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$ , then the length of the shortest line segment $P Q$ that is tangent to $C_{1}$ at $P$ and to $C_{2}$ at $Q$ is

15 units

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18 units

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20 units

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24 units

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Solution

The correct option is C $20 units$
Given circles are
$x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$
The centre and radius are
$C_{1} = (10, 0), r_{1} = 6 C_{2} = (- 15, 0), r_{2} = 9$
$C_{1} C_{2} = d = 25$

The transverse common tangent is the shortest common tangent, so the required length
$= \sqrt{d^{2} - (r_{1} + r_{2})^{2}} = \sqrt{(25)^{2} - (9 + 6)^{2}} = \sqrt{400} = 20 units$

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If C1 and C2 are circles whose equations are x2+y2−20x+64=0 and x2+y2+30x+144=0, then the length of the shortest line segment PQ that is tangent to C1 at P and to C2 at Q is

The correct option is C 20 unitsGiven circles are x2+y2−20x+64=0 and x2+y2+30x+144=0 The centre and radius are C1=(10,0), r1=6C2=(−15,0), r2=9 C1C2=d=25 The transverse common tangent is the shortest common tangent, so the required length =√d2−(r1+r2)2=√(25)2−(9+6)2=√400=20 units

If $C_{1}$ and $C_{2}$ are circles whose equations are $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$ , then the length of the shortest line segment $P Q$ that is tangent to $C_{1}$ at $P$ and to $C_{2}$ at $Q$ is