the pair of the tangent from the origin to the circlex^2+y^2+4x+2y+3=0 is
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Answered by
0
Answer:
Given circle x
2
+y
2
+4x+2y+3=0
Given point is origin (0,0)=(x
1
,y
1
)
So → S=x
2
+y
2
+4x+2y+3 ⇒(g,f)=(2,1)
S
1
=x
1
2
+y
1
2
+2gx
1
+2fy
1
+C
S
1
=0+0+2(2)0+2(1)0+3=3
Tangent,
T=xx
1
+yy
1
+g(x+x
1
)+f(y+y
1
)+C
=0x+0y+2(x+0)+1(y+0)+3
T=2x+y+3
Pair of tangents SS
1
=T
2
(x
2
+y
2
+4x+2y+3)3=(2x+y+3)
2
3(x
2
+y
2
)+12x+6y+9=(2x+y)
2
+3
2
+2.3(2x+y)
3(x
2
+y
2
)+12x+5y+9=(2x+y)
2
+12x+6y+9
3(x
2
+y
2
)=(2x+y)
2
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