If x2+y2+2x−4y+5=0, then the value of x+y is a) –1 b) 1 c) 2 d) ½
Answers
Answer:
ANSWER
Case I:-
Given that the line ax+by=0 touches the circle x
2
+y
2
+2x+4y=0, i.e., the line is tangent to the circle.
Equation of circle can be written as-
(x+1)
2
+(y+2)
2
−1−4=0
⇒(x+1)
2
+(y+2)
2
=(
5
)
2
Therefore,
Centre of circle =(−1,−2)
Radius of circle =
5
As we know that perpendicular distance from a point (x
1
,y
1
) to the line ax+by+c=0 is given by-
d=
a
2
+b
2
∣ax
1
+by
1
+c∣
∴⊥ distance from the centre of circle to the line ax+by=0-
d=
a
2
+b
2
∣−a−2b∣
As we know that the perpendicular distance from the centre of circle to the line touching the circle is equal to the radius of circle.
∴
a
2
+b
2
∣−a−2b∣
=
5
Squaring both sides, we have
a
2
+b
2
a
2
+4b
2
+4ab
=5
⇒a
2
+4b
2
+4ab=5a
2
+5b
2
⇒4a
2
+b
2
−4ab=0
⇒(2a−b)
2
=0
⇒2a−b=0.....(1)
Case II:-
Given that the line ax+by=0 is normal to the circle x
2
+y
2
−4x+2y−3=0, i.e., the line passes through the centre of circle.
The equation of circle can also be written as-
(x−2)
2
+(y+1)
2
−4−1−3=0
⇒(x−2)
2
+(y+1)
2
=(2
2
)
2
Centre of circle =(2,−1)
As the line passes through its centre, i.e., point (2,−1) will satisfy the equation ax+by=0.
∴2a−b=0.....(2)
From the given option, (C)(1,2) clearly satisfies the equation (1)&(2).
Hence the required answer is (C)(1,2).
Answer:
hope it will help you mark as brainlist answer plz
1and 2