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Let phi(x,y)=0 , be the equation of circ...

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

A

`(2,29/10)`

B

`(29/10,2)`

C

`(-2,(-29)/10)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
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)`
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