Verify that the circles
x2 + y2 – 4x – 7y + 6 = 0
x2 + y2 + 3x – 14y - 1 = 0 and
3(x2 + y2) + 2x – 35y + 4 = 0
have a common radical axis.
Answers
Answered by
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Step-by-step explanation:
:x
2
+y
2
−4x−6y+5=0
S
2
:x
2
+y
2
−2x−4y−1=0
S
3
:x
2
+y
2
−6x−2y=0
Now, the radical axis of the circles will be
S
1
−S
2
=0
(x
2
+y
2
−4x−6y+5)−(x
2
+y
2
−2x−4y−1)=0
2x+2y−6=0
⇒x+y=3.....(1)
Similarly,
S
2
−S
3
=0
(x
2
+y
2
−2x−4y−1)−(x
2
+y
2
−6x−2y)=0
2x−y=
2
1
.....(2)
S
3
−S
1
=0
(x
2
+y
2
−6x−2y)−(x
2
+y
2
−4x−6y+5)=0
2x−4y=−5.....(3)
Adding equation (1)&(2), we have
(x+y)+(2x−y)=3+
2
1
x=
6
7
Substituting the value of x in equation (1), we have
6
7
+y=3
⇒y=3−
6
7
=
6
11
Since the value of x and y also satisfies equation (3). Thus the radical centre of three circles is (
6
7
,
6
11
).
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