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Question

Let S1 be a circle passing through A(0,1) and B(2,2) & S2 is a circle of radius=10 such that AB is common chord of S1 & S2, find the equation of S2.

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Solution

Let, the equation of the circle S2 be
x2+y2+2gx+2fy+c=0,
whose centre is (g,f) and radius is g2+f2c.
According to the problem AB is also a chord of S2, so S2 also passes through A(0,1) and B(2,2).Then,
1+2f+c=0........(1) and
4+44g+4f+c=0........(2).
Subtracting (2) from (1) we get,
74g+2f=02(f+1)=4g5........(3).
Also, given radius of the circle S2 is 10.
Then,
g2+(f+1)2=10......(4). [Since the distance of the centre to A is the radius of S2.]
Now, using the value of (f+1) from (3) in (4) we get
g2+(4g5)24= 10
or, 4g2+16g240g+25=40
or, 20g240g15=0
or, 4g28g3=0
g=8±64+488=2±72.
Using the value of g in (3) we get,
f=3±272.
So, the equation of S2 be
(x+2±72)2+(y+3±272)2=10.


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