.How do I find the value of T, if the straightline y= T is the tangent to locus P with an equation of x^2 + y^2-2x+5y+5 =0?
Answers
Equation of circle =x
2
+y
2
−22x−4y+25=0
(x−11)
2
+(y−2)
2
=10
2
, radius 10 and r=(11,2)
Slope of line 5x+12y+8=0 is m
1
=
dx
dy
=
12
−5
So as we know two perpendicular has product of slope 2−1 i.e. m
1
m
2
2−1 where m
1
and m
2
are slope of two ⊥ lines
Here, if m
1
=
12
−5
then m
2
=
5
12
General equation of line with slope
5
12
⇒
y=
5
12
x+c.....(1)
perpendicular distance of tangent from center =10 (radius)
So,
(
5
12
)
2
+1
∣
∣
∣
∣
∣
5
12
(11)+C−2
∣
∣
∣
∣
∣
=10 as ⊥ distance =
a
2
+b
2
∣ax
1
+by
1
+c∣
from pt
∣5c+122∣=130
5c+122=130 5c+122=−130
C=
5
8
C=
5
−252
Put value of c in (1)
5y=12x+8 and 5y=12x−252 are lines ⊥ to 5x+12y+8=0 and are tangent to given will
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