it 3x - 2y +k=0 is tangent to the circle x2 +y2+6x-4y=0 then k=?
Answers
Answer:
Let us find points of intersection of line
3
x
+
4
y
−
k
=
0
and circle
x
2
+
y
2
=
16
. We can do this by putting value of
y
from first equation i.e.
y
=
k
−
3
x
4
and we get
x
2
+
(
k
−
3
x
)
2
16
=
16
or
16
x
2
+
k
2
+
9
x
2
−
6
k
x
=
256
i.e.
25
x
2
−
6
k
x
+
k
2
−
256
=
0
This would give two values of
x
and corresponding two values of
y
i.e. two points. But tangent cuts the circle in only at one point. This will be so when discriminant is zero i.e.
(
−
6
k
)
2
−
4
⋅
25
⋅
(
k
2
−
256
)
=
0
or
−
64
k
2
+
25600
=
0
or
k
=
±
20
graph{(x^2+y^2-16)(3x+4y-20)(3x+4y+20)=0 [-10, 10, -5, 5]}
Answer:
firstly we have to find out the points of intersection of line 3x+4y−k=0 and circle x2+y2=16. We can do this by putting value of y from first equation i.e. y=k−3x4 and we get
x2 + (k−3x)216 = 16
or 16x2+k2+9x2−6kx=256
i.e. 25 x 2−6kx + k2−256=0
This would give two values of x and corresponding two values of y i.e. two points. But tangent cuts the circle in only at one point. This will be so when discriminant is zero i.e.
(−6k)2−4⋅25⋅(k2−256)=0
or −64k2+25600=0 or k=±20