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Perpendicular Distance of a Point from a Line

If C1:x2+y2...

Question

If $C_{1} : x^{2} + y^{2} - 20 x + 64 = 0$ and $C_{2} : x^{2} + y^{2} + 30 x + 144 = 0$ . The line PQ touches the circle touches $C_{1}$ at $P$ and $C_{2}$ at $Q$ . Then the length of the shortest line segment $P Q$ is

20

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15

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22

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Solution

The correct option is A $20$
The equation of the first circle is $(x - 10)^{2} + y^{2} = 6^{2}$
The equation of the second circle is $(x + 15)^{2} + y^{2} = 9^{2}$
The distance between their centres $d$ is $10 - (- 15) = 25$
If length of traverse common tangent is $t_{1}$ , then $d^{2} = (r_{1} + r_{2})^{2} + t_{1}^{2}$
$\Rightarrow t_{1} = \sqrt{25^{2} - 15^{2}} = 20$
If length of direct common tangent is $t_{2}$ , then $d^{2} = (r_{2} - r_{1})^{2} + t_{2}^{2}$
$\Rightarrow t_{2} = \sqrt{25^{2} - 3^{2}} = \sqrt{616}$
$∴ t_{1} < t_{2} \Rightarrow P Q_{m i n .} = 20$

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Similar questions

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C_{2} : x^{2} + y^{2} + 30 x + 144 = 0

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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

If $C_{1} : x^{2} + y^{2} - 20 x + 64 = 0$ and $C_{2} : x^{2} + y^{2} + 30 x + 144 = 0$ . The line PQ touches the circle touches $C_{1}$ at $P$ and $C_{2}$ at $Q$ . Then the length of the shortest line segment $P Q$ is