Let `f(x,y)=0` be the equation of a circle. If `f(0,lamda)=0` has equal roots `lamda=2,2 and f(lamda,0)=0` has roots `lamda=(4)/(5),5` then the centre of the circle is

Options

A. `(2,29/10)`
B. `(29/10,2)`
C. `(-2,(-29)/10)`
D. None of these

Answer

B

Explanation

Let `phi(x,y)-=x^2+y^2+2gx+2fy+c=0`
`phi(0,lambda)-=0^2+lambda^2+0+2flambda+c=0` have equal roots , then 2+2=`-(2f)/1` and 2.2=`c/1`
`rArr ` f=-2 and c=4
And `phi(lambda, 0)=lambda^2+0+2g lambda)+0+c=0 " " therefore lambda^2+2glambda +c=0`
Here c=4 `therefore lambda^2 + 2glambda+4=0` have roots 4/5, 5
`therefore 4/5+5=-2g rArr g=-29/10` Centre `-=(-g,-f)=(29/10,2)`