A circle $S$ touches $Y$-axis at $(0,3)$ and makes an intercept of length 8 units on $\mathrm{X}$-axis. If…
A circle $S$ touches $Y$-axis at $(0,3)$ and makes an intercept of length 8 units on $\mathrm{X}$-axis. If the centre $\mathrm{C}$ of the circle $\mathrm{S}$ lies in the second quadrant, then the distance of $\mathrm{C}$ from the point $(-2,-1)$ is
$13$
$10$
$5$
$\sqrt{2}$
Solution
Equation of circle touches y-axis at $(0,3)$ is
$(x-a)^2+(y-3)^2=a^2 \Rightarrow x^2+y^2-2 a x-6 y+9=a^2$
Since the length of intercept on $\mathrm{x}$-axis at this circle is 8 unit
$\therefore 2 \sqrt{a^2-9}=8 \Rightarrow a^2=25 \Rightarrow a= \pm 5$
Center lies in second co-ordinate $\therefore a=-5$
So, the centre is $C(-5,3)$
Distance of the point $(-2,-1)$ from $C$ is
$\sqrt{(-2+5)^2+(-1-3)^2}=\sqrt{9+16}=5$