the guation of the circle passing t...
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the guation of the circle passing through the points (2,3) and (-1.1) and

JEE/Engineering Exams
Maths
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Let the equation of the required circle be ( (x-h)^{2}+(y-k)^{2}=r^{2} ) since the circle passes through points (2,3) and (-1,1) , ( (2-h)^{2}+(3-k)^{2}=r^{2} ldots(1) ) ( (-1-h)^{2}+(1-k)^{2}=r^{2} ldots(2) ) since the centre ( (h, k) ) of the circle lies on line ( x-3 y-11=0 ) ( h-3 k=11 ldots(3) ) From equations (1) and (2), we obtain ( (2-h)^{2}+(3-k)^{2}=(-1-h)^{2}+(1-k)^{2} ) ( 4-4 h+h^{2}+9-6 k+k^{2}=1+2 h+h^{2}+1-2 k+k^{2} ) ( 4-4 h+9-6 k=1+2 h+1-2 k ) ( 6 h+4 k=11 ldots(4) ) On solving equations (3) and (4), we obtain ( h=frac{7}{2} ) and ( k=frac{-5}{2} ) On substituting the values of ( h ) and ( k ) in equation ( (1), ) we obtain ( left(2-frac{7}{2}right)^{2}+left(3+frac{5}{2}right)^{2}=r^{2} ) ( Rightarrowleft(frac{4-7}{2}right)^{2}+left(frac{6+5}{2}right)^{2}=r^{2} ) ( Rightarrowleft(frac{-3}{2}right)^{2}+left(frac{11}{2}right)^{2}=r^{2} ) ( Rightarrow frac{9}{4}+frac{121}{4}=r^{2} ) ( Rightarrow frac{130}{4}=r^{2} )
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