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The equation of the tangents drawn from ...

The equation of the tangents drawn from the origin to the circle `x^(2)+y^(2)-2rx-2hy+h^(2)=0` are

A

x = 0

B

y = 0

C

`(h^(2)-r^(2))x-2rhy=0`

D

`(h^(2)-r^(2))x+2rhy=0`

Text Solution

Verified by Experts

Since, tangents are drawn from origin. So, the equation of tangent be y = mx
`rArr` Length of perpendicular from origin = radius

`rArr (|mr+h|)/(sqrt(m^(2+1)))=r`
`rArr m^(2)r^(2)+h^(2)+2mrh=r^(2)(m^(2)+1)`
`rArr m=|(r^(2)-h^(2))/(2rh)|,m=oo`
`therefore" Equation oftangents are " y=|(r^(2)-h^(2))/(2rh)|x,x=0` Therefore (a) and (c) are the correct answers.
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