Question

Let $f\left(x\right)=0$ be the equation of a circle, if $f(0,k)=0$ has equal roots $k=2,2$ and $f(k,0)=0$ roots $k=\frac{4}{5},5$, then the centre of the circle is ?

Answer 1

Let $f(x,y)=x^2+y^2+2gx+2fy+c=0$ be the equation of the circle.

$f(0,k)=k^2+2fk+c=0$

At $k=2$

$\Rightarrow 4+2f\left(2\right)+c=0$

$\Rightarrow 4+4f+c=0\to \left(1\right)$

$f(k,0)=k^2+0+2gk+c=0$

At $k=\frac{4}{5}$

$\Rightarrow \frac{16}{25}+\frac{8g}{5}+c=0$

$\Rightarrow 16+40g+25c=0\to \left(2\right)$

At $k=5$

$25+10g+c=0$

Solving $\left(1\right)$ and $\left(3\right)$

$\Rightarrow 4+4f=25+10g$

$\Rightarrow 10g-4f=-21\to \left(4\right)$

Solving $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 16+40g+25(-4-4f)=0$

$\Rightarrow 16+40g-100-100f=0$

$\Rightarrow 40g-100f=84$

$\Rightarrow 10g-25f=21\to \left(5\right)$

Solving $\left(4\right)$ and $\left(5\right)$

$10g-4f=-21-(10g-25f=21)\to 21f=-42$

$\Rightarrow f=-2$

$c=-4f-4=8-4=4$

$g=\frac{1}{10}(4f-21)$

$=\frac{1}{10}(-8-21)$

$=\frac{-29}{10}$

Centre $\to (\frac{29}{10},2)$