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Question: Equation of the pair of tangents drawn from the origin to the circle \(x ^ { 2 } + y ^ { 2 } + 2 g x...

Equation of the pair of tangents drawn from the origin to the circle x2+y2+2gx+2fy+c=0x ^ { 2 } + y ^ { 2 } + 2 g x + 2 f y + c = 0is.

A

gx+fy+c(x2+y2)g x + f y + c \left( x ^ { 2 } + y ^ { 2 } \right)

B

(gx+fy)2=x2+y2( g x + f y ) ^ { 2 } = x ^ { 2 } + y ^ { 2 }

C

(gx+fy)2=c2(x2+y2)( g x + f y ) ^ { 2 } = c ^ { 2 } \left( x ^ { 2 } + y ^ { 2 } \right)

D

(gx+fy)2=c(x2+y2)( g x + f y ) ^ { 2 } = c \left( x ^ { 2 } + y ^ { 2 } \right)

Answer

(gx+fy)2=c(x2+y2)( g x + f y ) ^ { 2 } = c \left( x ^ { 2 } + y ^ { 2 } \right)

Explanation

Solution

Equation of pair of tangents is SS1=T2S S _ { 1 } = T ^ { 2 },

Where T=xx1+yy1+g(x+x1)+f(y+y1)+cT = x x _ { 1 } + y y _ { 1 } + g \left( x + x _ { 1 } \right) + f \left( y + y _ { 1 } \right) + c

⇒c(x2+y2+2gx+2fy+c)=(gx+fy+c)2\Rightarrow c \left( x ^ { 2 } + y ^ { 2 } + 2 g x + 2 f y + c \right) = ( g x + f y + c ) ^ { 2 }

⇒c(x2+y2)=(gx+fy)2\Rightarrow c \left( x ^ { 2 } + y ^ { 2 } \right) = ( g x + f y ) ^ { 2 }.