Question

If $l,m,n$ denote the lengths of the intercepts made by the circle $x^2+y^2-8x+10y+16=0$ on $x$-axis, $y$-axis and $y=-x$ respectively, then last digit of odd prime factor of $l^2+10m^2+26n^2$ is equal to

Answer 1

Given equation of circle is
$x^2+y^2-8x+10y+16=0$
Here, $g=-4,f=5,c=16$
Now, length of intercept made by circle on x-axis is $l=2\sqrt{g^2-c}$
$\Rightarrow l=2\sqrt{16-16}=0$
Length of intercept made by circle on y-axis is $m=2\sqrt{f^2-c}$
$\Rightarrow m=2\sqrt{25-16}=6$
Length of intercept made by the circle on the line $y=-x$
So, point of intersection of circle and the line are $(8,-8)$ and $(1,-1)$
Length of intercept made by circle on the line is $n=\sqrt{7^2+(-7{)}^2}=7\sqrt{2}$

$\Rightarrow l=0,m=6,n=7\sqrt{2}$

$\Rightarrow l^2+10m^2+26n^2=360+2548=2908$

So, the odd prime factor is $727$ and its last digit is $7$