Question

show that the tangent at (-1,2) of the circle x2+y2-4x-8x+7=0 touches the circles x2+y2+4x+6y=0 and also find the point of contact
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Answer 1

Step-by-step explanation:

The given circles be

$S \colon {x}^{2} + {y}^{2} - 4x - 8y + 7 = 0 \\ S^{ \prime} \colon {x}^{2} + {y}^{2} + 4x + 6y = 0$

The equation of tangent at $(-1,2)$is $T=0$ i.e.,

$T \colon ( - 1)x + (2)y - 2( - 1 + x) - 4(2 + y) + 7 = 0$

$T \colon - x + 2y + 2 - 2 x - 8 - 4 y + 7 = 0$

$T \colon - 3x - 2y + 1 = 0$

$T \colon 3x + 2y - 1 = 0$

Now, centre of S' be C' and radius be r'

so,

$C^{ \prime} \equiv( - 2, - 3)$

$r^{ \prime } = \sqrt{4 + 9} = \sqrt{13}$

Perpendicular distance from (-2,-3) to T=0 will be the radius,

so,

$d = \frac{ | - 6 - 6 - 1| }{ \sqrt{9 + 4} } = \frac{13}{ \sqrt{13} } = \sqrt{13}$

Since, d=r'

hence 3x+2y-1=0 touches S'=0

Answer 2

Answer:

A violin string vibrates with fundamental frequency of 220 Hz. The frequency of first overtone is A 440 Hz (B) 420 Hz © 410 Hz D 400 Hz

Step-by-step explanation:

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