Question

Find the equation to the circle which touches the x-axis and passes through the two points(1, -2) and (3, -4).

Answer 1

For the circle

$x^2+y^2+2gx+2fy+c=0$

From the figure $f=r=\sqrt{g^2+f^2-c}$

$\Rightarrow g^2=c$

And point $(1,-2),(3,-4)$ are on the circle

Then, we have

$5+2g-4f+g^2=0$ .........(1)

$25+6g-8f+g^2=0$ .......(2)

From (1) and (2)

$g^2-2g-15=0$

$g=-5,g=3$

$g=-5$ rejected because the circle lies in the 4th quadrant

So, $g=3,f=2,c=9$

Equation of circle becomes

$x^2+y^2-6x+4y+9=0$