Question

Suppose the equation of the circle which touches both the coordinates and passes through the point with abscissa $-2$ and ordinate $1$ has the equation $x^2+y^2+Ax+By+C=0$, find all the possible ordered triplet $(A,B,C)$.

Answer 1

REF. Image.

From $x^2+y^2+Ax+By+c=0$

coordinates of centre (0) $=(\frac{-A}{2},\frac{-B}{2})$

Radius = $\sqrt{\frac{A^2}{4}+\frac{B^2}{4}-C}\Rightarrow (B<0)$

$OA=\frac{A}{2},OB=\frac{-B}{2}$

As $OA=OB,\Rightarrow A=-B\Rightarrow A+B=0$

& OA = Radius

$\Rightarrow \frac{A^2}{4}+\frac{B^2}{4}-C=\frac{A^2}{4}$

$\Rightarrow \frac{B^2}{4}=C$

$\Rightarrow$ circle's equation :

$x^2+y^2+Ax-Ay+\frac{A^2}{4}=0$

$(-2,1)4+1-2A-A+\frac{A^2}{4}=0$

$\Rightarrow A^2-12A+20=0$

$A^2-10A-2A+20=0$

$\Rightarrow A(A-10)-2(A-10)=0\Rightarrow (A-2)(A-10)=0$

$\Rightarrow A=2$ or $A=10$

$\Rightarrow B=-2$ or $B=-10$

$\Rightarrow C=1$ or $C=25$

$\Rightarrow$ 2 possible triplets.