Question

The intercept made by the circle $x^2+y^2-4x-6y-3=0$ on the line $x+y-3=0$ is

• A. $\sqrt{14}$
• B. $2\sqrt{14}$
• C. $6\sqrt{2}$
• D. $3\sqrt{2}$

Answer 1

The correct option is B $2\sqrt{14}$

The general equation of cirlce is $x^2+y^2+2gx+2fy+c=0$ where

Center = $(-g,-f)$

Radius = $\sqrt{g^2+f^2-c}$

So, for given circle,

$Center=(2,3)$

$Radius=\sqrt{2^2+3^2+3}=4$

The distance of point (h,k) from the line lx + my + n = 0 is given by

$|\frac{lh+mk+n}{\sqrt{l^2+m^2}}|$

So, distance of chord from center,

$=\frac{2+3-3}{\sqrt{1^2+1^2}}=\sqrt{2}$

Now, in triangle OAB,

$A{B}^2+B{O}^2=A{O}^2$

$\to AB=\sqrt{4^2-(\sqrt{2}{)}^2}=\sqrt{14}$

So, intercept formed by circle on line x+y-3 =0

$=2AB$

$=2\sqrt{14}$

Hence, (B) is the correct answer.