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  • Let be the equation of a circle. If <img alt="f(0, \lambda)=0" src="https://entrancecorner.oncodeco

Let f(x, y)=0 be the equation of a circle. If f(0, \lambda)=0 has equal roots \lambda=2,2 and f(\lambda, 0)=0 has roots \lambda=\frac{4}{5}, 5, then the centre of the circle is

Option: 1

\left(2, \frac{29}{10}\right)


Option: 2

\left(\frac{29}{10}, 2\right)


Option: 3

\left(-2, \frac{29}{10}\right)


Option: 4

\mathrm{None \, \, of \, \, these}


Answers (1)

\text { \ } \quad f(x, y)=x^2+y^2+2 g x+2 f y+c=0

Now, f(0, \lambda) \equiv \lambda^2+2 f \lambda+c=0 and its roots are 2,2 \text {. }   \therefore 2+2=-2 f, 2 \times 2=c \text {, i.e. } f=-2, c=4

f(\lambda, 0) \equiv \lambda^2+2 g \lambda+c=0 and its roots are \frac{4}{5}, 5 .

\therefore \frac{4}{5}+5=-2 g, \frac{4}{5} \times 5=c, \quad \text { i.e., } g=\frac{-29}{10}, c=4.

Hence, centre of the circle =(-g,-f)=\left(\frac{29}{10}, 2\right).

Posted by

Ramraj Saini

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