Question

If a circle passes through the points of intersection of the axes with the lines $\lambda x-y+1=0$ and $x-2y+3=0,\lambda >0$ and radius of the circle is $\frac{k}{l\sqrt{2}}$ then value of $(k+l)$ is : ($k$ and $l$ are co-prime)

Answer 1

$\lambda x-y+1=0$ intersects the axes at A$(-\frac{1}{\lambda },0)$ and B$(0,1)$

$x-2y+3=0$ intersects the axes at C$(0,\frac{3}{2})$ and D$(-3,0)$

Let center of circle be at O$(h,k)$

Then OA = OB = OC = OD

So $(h-0{)}^2+(k-1{)}^2=(h-0{)}^2+(k-\frac{3}{2}{)}^2$

$\Rightarrow 1-2k=\frac{9}{4}-3k$

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

$(h-0{)}^2+(k-1{)}^2=(h-(-3){)}^2+(k-0{)}^2$

$1-2k=9+6h$

$h=-\frac{7}{4}$

So centre is at O$(\frac{-7}{4},\frac{5}{4})$

Hence by distance formula

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

OB $=\sqrt{\frac{50}{4}}$

$=\frac{\sqrt{50}}{2}$

$=\frac{5\sqrt{2}}{2}$

$=\frac{5}{\sqrt{2}}$

$1$ and $5$ are co-prime. Required answer is $5+1=6$.