Question

The length of chord of contact of the point $(3,6)$ with respect to the circle $x^2+y^2=10$ is

• A. $\frac{2\sqrt{70}}{3}$
• B. $6\sqrt{5}$
• C. $\sqrt{5}$
• D. $\frac{12}{\sqrt{5}}$

Answer 1

Answer: (A)

$\frac{2\sqrt{70}}{3}$

The chord of contact of the tangents from a point $\left(x_1{y}_1\right)$ to the circle $x^2+y^2=a^2$ is given by

${xx}_1+{yy}_1=a^2$

So the equation of the chord of contact of the point $(3,6)$ with respect to the circle $x^2+y^2=10$ is

$3x+6y=10⟶\left(1\right)$

the perpendicular distance of the line from the origin $=OA=\left|\frac{3\times 0+6\times 0-10}{\sqrt{3^2+6^2}}\right|=\frac{10}{\sqrt{45}}$

and $OB=\sqrt{10}$

Using pythagoras theorem is $\Delta OAB$ we can write

${OB}^2={OA}^2+{AB}^2$

$\Rightarrow {AB}^2=10-\frac{100}{45}=\frac{350}{45}$

$\Rightarrow AB=\frac{\sqrt{10}}{3}$

$\therefore$ length of the chord $=2AB=\frac{2\sqrt{70}}{3}$

Answer : Option A.