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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

If C1: x2+y...

Question

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

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Solution

The correct option is A $10$
$C 1 : x^{2} + y^{2} - 20 x + 64 = 0$

$⟹ (x - 10)^{2} + y^{2} = 6^{2}$

$C 1 = (10, 0), r 1 = 6$

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

$⟹ (x + 15)^{2} + y^{2} = 9^{2}$

$C_{2} = (- 15, 0), r_{2} = 9$

Shortest distance = $C 1 C 2 - r_{1} - r_{2}$

$⟹ 25 - 6 - 9 = 25 - 15 = 10$

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If C1:x2+y2−20x+64=0 and C2:x2+y2+30x+144=0. Then the length of the shortest line segment PQ which touches C1 at P and to C2 at Q is ?

The correct option is A 10C1:x2+y2−20x+64=0 ⟹(x−10)2+y2=62 C1=(10,0),r1=6 C2:x2+y2+30x+144=0 ⟹(x+15)2+y2=92 C2=(−15,0),r2=9 Shortest distance = C1C2–r1−r2 ⟹25–6–9=25–15=10

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