Find the equation of the circle which passes through the point (4,1) (6,5)and having the centre on the line 4x+3y-24=0
Answers
Step-by-step explanation:
Since C(α,β) be the centre of the circle.
Since the circle passes through the point (2,8)
∴ Radius of the circle=
(α−2)
2
+(β−8)
2
Since the circle touches the lines 4x−3y−24=0 and 4x+3y−42=0
∴
∣
∣
∣
∣
∣
4
2
+5
2
4α−3β−24
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
4
2
+5
2
4α+3β−42
∣
∣
∣
∣
∣
=
(α−2)
2
+(β−8)
2
........(1)
Solving them, we get
4α−3β−24=±(4α+3β−42) ........(3)
∴6β=18 or β=3 by taking the positive sign
and 8α=66 or α=
8
66
=
4
33
by taking negative sign.
Given ∣α∣≤8⇒−8≤α≤8
∴α
=
4
33
Put β=3 in equations (1) and (3) and equating, we get
(4α−33)
2
=(α−2)
2
+25
⇒16α
2
−264α+1089=25α
2
+725−100α
⇒9α
2
+164α−364=0 which is quadratic in α
∴α=
2×9
−164±
164
2
−36×−364
=
18
−164±
164
2
+36×364
=
18
−164±
40000
=
18
−164±200
=2,
9
−182
But −8≤α≤8 ∴α=2
Now r
2
=(α−2)
2
+(3−8)
2
=(2−2)
2
+(3−8)
2
=25 where r is the radius of the circle
Hence the required equation of the required circle is
(x−2)
2
+(y−3)
2
=25
x
2
+y
2
−4x−6y+4+9−25=0 or x
2
+y
2
−4x−6y−12=0