fine the equation of the pair of transverse common tangents of x²+y²-4x-20y+28=0,x²+y²+4x-6y+4=0.
Answers
Answer:
Step-by-step explanation:
x
2
+y
2
−4x−10y+28=0 x
2
+y
2
+4x−6y+4=0
centre c
1
=(2,5) centre c
2
=(−2,3)
r
1
=
4+25−28
r
2
=
4+9−4
r
1
=1 r
2
=3
Distance between c
1
and c
2
=
(2+2)
2
+(5−3)
2
=
20
∴c
1
c
2
>r
1
+r 2 [ 2 >4]
∴ Point P divides c
1 c 2 in the ratio 1:3
P = [2+ 4 (−2−2) ,5+4 (3−5] ]
P = (1, 2 ,9 )
Eqn of tangent with slope m is
y− 2 9
=m(x−1) ⇒ 2mx−2y+9−2m=0
The above line is tangent to circle x
2 +y 2 −4x−10y+28=0
Radius = bot distance from centre (2,5) to line
1 =4m +4
2m(2)−2(5)+9−2m
⇒ 1 =
∣
4m
2
+4
2m−1
∣
∣
∣
∣
∣
4m
2
+4=4m
2
+1−4m
4m+3=0 ⇒
m=−
4
3
Since we have two tangents passing from point (1,
2
9
)
But m
2
term is eliminated
∴ slope of other line is ∞
Eqn. of tangent having slope −
4
3
is
y−
2
9
=−
4
3
(x−1) ⇒
3x+4y−21=0
eqn. of tangent having slope ∞ & passing though (1,
2
9
) is
x=1
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