Question

Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1.
Show that there are two such circles.

Answer 1

Let $x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
be the required circle
Now (i) passes through P (1, I) and (2, 2)
$\therefore 1+1+2g+2f+c=0\dots \left(ii\right)4+4+4g+4f+c=0\dots \left(iii\right)$
Also radius = 1
$\Rightarrow \sqrt{g^2+f^2-c}=1\Rightarrow g^2+f^2-c=1$
from (ii) and (iii)
$g+f+\frac{c}{2}=-1$ and
$g+f+\frac{c}{4}=-2$
On subtracting
$c=4$
and $g+f=—3\dots \left(v\right)$
from (iv) $g^2+f^2=5$
$[\because (g+f{)}^{2}g2+f^2+2gf\Rightarrow 9=5+2gf]\therefore 2gf=4gf=2$
so, $(g-f{)}^2=(g+f{)}^2-4gf=9-8=1$
$f=-2$ or $-1$
Thus, required circle
$x^2+y^2-2x-4y+4=0,x^2+y^2-4x-2y+4=0$