Dear Devashish,

Hi, 
Generally speaking the center of the circle which equation is 
(x-a)^2+(y-b)^2=R^2 is the point M(a,b) and the radius of this circle is R 
(x – 2)^2 + (y + 3)^2 = 17 the center of this circle is A(2,-3)and the center of the circle (x – 5)^2 + y^2 = 32 is B(5,0)
The line (D): y=ax+b which passes through A and B is such that
A belongs to (D) ==> -3=a(2)+b=2a+b ==> b=-2a-3(*)
B belongs to (D) ==> 0=a(5)+b=5a+b ==>b=-5a(**)
(*) and (**) lead to -5a=-2a-3 ==> 5a-2a-3=0 ==> 3a-3=0 ==>3a=3
==> a=3/3=1 ==> a=1 From (**) b=-5a=-5(1)=-5
Finally the equation of (D) is y=x-5

The circle (C) with the equation (x-a)^2+(y-b)^2=r^2 has the same center with the circle which equation is (x – 4)^2 +(y + 3)^2 = 49 means a=4 and b=-3 O (4,-3) is the center of the circle (C). Therefore the equation of (C) (x – 4)^2 +(y + 3)^2=r^2 where r is the radius of (C) If (C) is tangent to the y-axis this (C) intersects the y-axis( which equation is x=0) at only one point M(x'', y'') x''=0 because M belongs to the y-axis M belongs to (C) y''=-3 because M is the orthogonal projection of O on the y-axis. The radius of (C) is the distance between O and M r=OM=sqrt[(x''-4)^2+(y''+3)^2]=sqrt[(0-4)^…
=sqrt[(-4)^2+(0)^2]=sqrt16=4 Finally the radius of (C) is r=4
Finally the equation of (C) is (x-4)^2+(y+3)^2=4^2=16

The area of the circle (C) is A =pi*r^2 where pi=3.14 and r is the radius of (C) A=3,14(4)^2=50.24

The area A'' of the circle (x-4)^2+(y+3)^2=49=7^2 is given by A''=pi*r''^2 where r'' is its radius . From the equation above we draw r=7 then A''=3.14(7)^2=153.86. The area S of the ring formed by the two circles is S= A''-A=7^2*pi-4^2*pi= 49pi-16pi=33pi units squared 

The line x=9 is secant to the circle (x – 6)^2 + (y + 2)^2 = 9 if by replacing x by its value in the equation of the circle we find two values of y i.e the equation (9-6)^2+(y+2)^2=9 has two roots then we will deduct that the line and the circle are secant (9-6)^2+(y+2)^2=9 ==> (3)^2 +(y+2)^2=9 
==> 9+(y+2)^2=9 => (y+2)^2=9-9=0 ==> (y+2)^2=0 ==> y+2=0
y=-2 One value for y therefore we can deduct that the line x=9 is tangent to the circle (x – 6)^2 + (y + 2)^2 = 9. If we had not found any value for y we would have deducted that the line and the circle are neither secant nor tangent

The circle (C) (x-a)^2+(y-b)^2=r^2 contain the points J(-6,0) K(-3,3) and L(0,0) 
J belongs to (C) ==> (-6-a)^2+(0-b)^2=r^2 ==> (-6-a)^2+(-b)^2=r^2 (*)
K belongs to (C) ==> (-3-a)^2+(3-b)^2=r^2 (**)
L belongs to (C) ==> (0-a)^2+(0-b)^2=r^2 ==> (-a)^2+(-b)^2=r^2 ==>
a^2+b^2=r^2 (***)

(*) ==> (-6-a)^2+(-b)^2=r^2 ==> 36-12a+a^2+b^2=r^2 ==> 
36-12a=-a^2-b^2+r^2=0 because from (***) a^2+b^2=r^2
Hence 36-12a=0 ==> 12a=36 ==> a=36/12=3 ==> a=3

(**) ==> (-3-a)^2+(3-b)^2=r^2 ==> 9-6a+a^2+9-6b+b^2=r^2 
==> 18-6a-6b+a^2+b^2=r^2 ==> 18-6a-6b=-a^2-b^2+r^2=0 
==> 18-6a-6b=0 By replacing a by its value we obtain 18-6(3)-6b=0
==. 18-18-6b=0 ==> -6b=0 ==> b=0
a^2+b^2=r^2 ==> 3^2+0^2=r^2 ==> 9=r^2 ==> r=-3 or r=3. Since r>=0 we deduct r=3
Finally the equation of the circle (C) is (x-3)^2+(y-0)^2=3^2
==> (x-3)^2+y^2=3^2

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert