Question

The centre of the circle touching $y-$axis at $(0,4)$ and making an intercept $2$ units on the positive $x-$axis is

• A. $(4,\sqrt{3})$
• B. $(\sqrt{17},3)$
• C. $(\sqrt{17},4)$
• D. $(4,\sqrt{17})$

Answer 1

Answer: (C)

$(\sqrt{17},4)$

Circle touches $y-$axis at $(0,4)$ means $y-$axis is tangent to circle, and $y=4$ line passes through centre of circle.

$\therefore$ Ordinate of centre of circle $=4$ & radius$=x-$coordinate of centre.

Centre $=(h,4)$

$\therefore (x-h{)}^2+(y-4{)}^2=h^2$

And given $x-$intercept$=2$

Put $y=0$ in the above equation of circle

$(x-h{)}^2+(0-4{)}^2=h^2$
$\Rightarrow (x-h{)}^2+16=h^2$
$\Rightarrow x^2-2hx+16=0$

Sum of roots: $x_1+x_2=2h$
Product of roots: $x_1{x}_2=16$
$\therefore (x_1-x_2)=\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}$
$\Rightarrow 2=\sqrt{4h^2-4\times 16}$
$\Rightarrow 4=4h^2-64$
$\Rightarrow h^2=\frac{68}{4}$
$\Rightarrow h=\sqrt{17}$

$\therefore$ Centre of circle $=(\sqrt{17},4)$