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Question

Find the equation of the circle passing through the points (2,3) and (-1,1) and whose centre is on the line x3y11=0.
 


Solution

The equation of the circle is 
(xh)2+(yk)2=r2
Since the circle passes through point (2,3)
(2h)2+(3k)2=r24+h24h+9k26k=r2h2+k24h6k+13=r2
  Also the circle passes through piont (-1,1)
(1h)2+(1k)=r21+h2+2h+1+k22k=r2h2+k2+2h2k+2=r2
From (ii) and (iii) we have
h2+k24h6k+13=h2+k2+2h2k+26h4k=116h+4k=11
 Since the centre (h, k) of the circle lies on the line x3y11=0
 h3k11=0h3k=11
 Solving (iv) and (v) we have
h=72 and k=52
Putting these value of h and k in (ii), we have
(72)2+(52)24×726×52+13=r3
494+25414+15+13=r2r2=652
 Thus equation of required circle is
(x72)2+(y+52)2=652x2+4947x+y2+254+5y=652x2+4947x+y2+254+5y=6524x2+4928x+20y56=04(x2+y27x+5y14)=0x2+y27x+5y14=0.
 

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