If power of a point $(4,2)$ with respect to the circle $x^2+y^2$ $-2 \alpha x+6 y+\alpha^2-16=0$ is 9 , then…
If power of a point $(4,2)$ with respect to the circle $x^2+y^2$ $-2 \alpha x+6 y+\alpha^2-16=0$ is 9 , then the sum of the lengths of all possible intercepts made by such circles on the coordinate axes is
$16+4 \sqrt{6}$
$16+4 \sqrt{6}-6 \sqrt{2}$
$16+4 \sqrt{6}+6 \sqrt{2}$
$16+6 \sqrt{2}$
Solution
$x^2+y^2-2 \alpha x+6 y+\alpha^2-16=0$
Centre $(C)=(\alpha,-3) ;(x-\alpha)^2+(y+3)^2=25$
Radius $(R)=5$
$d=$ distance between centre and $(4,2)$
$\Rightarrow d^2=(\alpha-4)^2+25$
Power $=d^2-R^2 \Rightarrow 9=(\alpha-4)^2 \mathrm{~m} \Rightarrow \alpha=1,7$
When $\alpha=1:(x-1)^2+(y+3)^2=25$
Put $x=0 \Rightarrow(y+3)^2=24$
$\Rightarrow y^2+6 y-15=0 \Rightarrow y=-3 \pm 2 \sqrt{6}$
Put $y=0 \Rightarrow(x-1)^2=16 \Rightarrow x=5,-3$
When $\alpha=7:(x-7)^2+(y+3)^2=25$
put $y=0 \Rightarrow(x-7)^2=16 \Rightarrow x=11,3$
Sum of length of intercept $=16+4 \sqrt{6}$