Question

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

• A. $x^2+y^2-10x+10y+25=0$ and $(-10,10,25)$
• B. $x^2+y^2+2x-2y+1=0$ and $(2,-2,1)$
• C. $x^2+y^2-2x+2y+1=0$ and $(-2,2,1)$
• D. $x^2+y^2+10x-10y+25=0$ and $(10,-10,25)$

Answer 1

Answer: (B), (D)

$x^2+y^2+10x-10y+25=0$ and $(10,-10,25)$
$x^2+y^2+2x-2y+1=0$ and $(2,-2,1)$

Let the centre of the circle be $(-r,r)$ where $r$ is the radius of the circle
$\Rightarrow$ equation of circle will be:
${(x+r)}^2+{(y-r)}^2=r^2$
$\Rightarrow$ $x^2+2rx+r^2+y^2-2ry+r^2=r^2$
$\Rightarrow$ $x^2+y^2+2rx-2ry+r^2=0$
passes through $(-2,1)$
$\Rightarrow$ $r^2-6r+5=0$ $\to$ $r=1,5$
when $r=1,x^2+2x+y^2-2y+1=0$
Hence $A=2,B=-2,C=1$
Also when $r=5$
$x^2+10x+y^2-1oy+25=0$
$\Rightarrow$ $A=10,B=-10,C=25$
Hence the required triplets are $(2,-2,1)$ and $(10,-10,25)$
$-\frac{A}{2}=-1,-5$ $\Rightarrow$ $B=2,-10$
Also $\sqrt{g^2+f^2-c}=r$
$\Rightarrow$ $\frac{A^2}{4}+\frac{B^2}{4}-r^2=C$ $\Rightarrow$ $C=1,25$