Question

Chord is drawn to the circle $x^2+y^2-4x-2y=0$ at a point where it cuts the x-axis whose slope is parallel to the tangent at an origin. The intercept of the chord on y-axis is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (D)

$8$

$S:(x-2{)}^2+(y-1{)}^2=5$

Let $y–1=m(x–2)+\sqrt{5}\sqrt{m^2+1}$ be any tangent to S.

Tangent at $(0,0)$ is given by slope of

$0–1=m(0–2)+\sqrt{5}\sqrt{m^2+1}$

$(2m-1{)}^2=5(m^2+1)$

$⟹m^2+4+4m=0$

$⟹m=-2$

The circle cuts the $x$ – axis at $y=0$

$⟹(x–2{)}^2=4$

$⟹x=4$

Equation of chord which passes through $(4,0)$ and has slope $=-2$ is

$\frac{y–0}{x–4}=-2⟹2x+y=8$

$⟹\frac{x}{4}+\frac{y}{8}=1$

Or

$2\left(0\right)+y=8⟹y=8$

Intercept on $y$ – axis by the chord is $8$.