Question

The length of the chord of contact from point $(2,1)$ to the circle $x^2+y^2+4x+2y+1=0$ is

• A. $\frac{8}{\sqrt{5}}$
• B. $\frac{4}{\sqrt{5}}$
• C. $\frac{4\sqrt{6}}{\sqrt{5}}$
• D. $\frac{2\sqrt{6}}{\sqrt{5}}$

Hint:

Chord of Contact.
For external point, use $T=0$ to find line of chord. Then compute perpendicular from center to line, and apply $2\sqrtR^2 - d^2$.

Answer 1

The Correct Option is A

Chord of contact from $(x_1,y_1)$ to circle $S$: \[ T = 0: xx_1 + yy_1 + g(x + x_1) + h(y + y_1) + c = 0 \] From $S$: $g = 2$, $h = 1$, $c = 1$, and $(x_1, y_1) = (2,1)$: \[ T: 2x + y + 4 + 2 + 1 = 0 \Rightarrow 2x + y + 3 = 0 \] Distance from center $(-2, -1)$ to chord: \[ d = \frac{|2(-2) + (-1) + 3|}{\sqrt{5}} = \frac{2}{\sqrt{5}} \] \[ R = \sqrt{g^2 + h^2 - c} = \sqrt{4 + 1 - 1} = 2 \] Chord length = $2\sqrt{R^2 - d^2} = 2\sqrt{4 - \frac{4}{5}} = \frac{8}{\sqrt{5}}$