Question

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

• A. $(2,\frac{29}{10})$
• B. $(\frac{29}{10},2)$
• C. $(-2,\frac{29}{10})$
• D. $(\frac{29}{10},-2)$

Answer 1

The correct option is B $(\frac{29}{10},2)$

$f(x,y)=x^2+y^2+2gx+2fy+c=0$
Now,
$f(0,\lambda )=0\Rightarrow {\lambda }^2+2f\lambda +c=0$
Which has its roots as $2,2$
$2+2=-2f\Rightarrow f=-22\times 2=c\Rightarrow c=4$

$f(\lambda ,0)=0\Rightarrow {\lambda }^2+2g\lambda +c=0$
Which has its roots as $\frac{4}{5},5$
$\frac{4}{5}+5=-2g\Rightarrow g=-\frac{29}{10}\frac{4}{5}\times 5=c\Rightarrow c=4$

Hence, the centre of circle is $(-g,-f)=(\frac{29}{10},2)$



Alternate solution:
Quadratic in $y$ with roots $2,2$ is
$y^2-4y+4=0\cdots \left(1\right)$
Quadratic in $x$ with roots $\frac{4}{5},5$ is
$(x-\frac{4}{5})(x-5)=0\Rightarrow x^2-\frac{29}{5}x+4=0\cdots \left(2\right)$
Therefore, the equation of circle is
$x^2+y^2-\frac{29}{5}x-4y+8=0$
So, the centre of the cirlce is
$C=(-g,-f)=(\frac{29}{10},2)$