Question

The length of the intercept made by the normal at (1,6) of the circle $x^2+y^2-4x-6y+3$ = 0 between the Coordinate axis is

• A. $\sqrt{5}$
• B. $\sqrt{5}$
• C. $3\sqrt{10}$
• D. $\sqrt{\frac{5}{2}}$

Answer 1

The correct option is

C

$3\sqrt{10}$

Given point P(1,6) on the circle $x^2+y^2-4x-6y+3$ = 0 and comparing it with $x^2+y^2+2hx+2ky+c=0$
$\Rightarrow h=-2,k=-3$
Thus, the centre $C(-h,-k)=(2,3)$

Equation of line joining CP, $\frac{y-3}{6-3}$ = $\frac{x-2}{1-2}$

$\Rightarrow \frac{y-3}{3}$ = $\frac{x-2}{-1}$

$\Rightarrow -y+3$ = $3x-6$

$\Rightarrow 3x+y-9$ = 0 $\Rightarrow \frac{x}{3}+\frac{y}{9}$ = 1 is the required normal equation.
Whose x-intercept is $3$ and y-intercept is $9$.

So, the length of Intercept made by the normal = $\sqrt{9+81}$ = $\sqrt{90}$ = $3\sqrt{10}$