14. The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11,9) and has diameter 10/2 units.
Answers
LetthegivenpointP(x
1
,y
1
)=(11,−9)lieonacirclewithcentreat
C(x
2
,y
2
)=(2a,a−7)&radius=r.
ThenPC=r..(i)
Giventhatthediameterofthecircle=10
2
units.
∴r=
2
diameter
=
2
10
2
units
=5
2
units..(ii).
Again,bydistanceformula,d=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
∴PC=d=
(11−2a)
2
+(−9−a+7)
2
=
5a
2
−40a+125
.
∴From(i)&(ii)weget
5a
2
−40a+125=(5
2
)
2
⟹5a
2
−40a+75=0
⟹(a−5)(5a−3)=0
⟹a=5,
5
3
Ansa=5,
5
3
Answer:
Below
Step-by-step explanation:
D= 5
Radius= 5/2
Equation of circle is (x-a)^2 + (y-b)^2 = r^2
It passes through pts. (11,9). So, these points must satisfy equation of circle
Centre of circle is (2a,a-7)
x=11 y=9 a=2a b= a-7
Therefore, (11-2a)^2 +[9-(a-7)]^2 = (5/2)^2
121 +4a^2 - 44a + (16 - a) ^2 = 25/4
4[121 + 4a^2 - 44a + 256 + a^2 + 32a] = 25
4[378 + 5a^2 - 12a] =25
On further solving you will get the value of A