Question

Let the tangents drawn from the origin to the circle, $x^2+y^2-8x-4y+16=0$ touch it at the points $A$ and $B$. The $(AB{)}^2$ is equal to :

• A. $\frac{32}{5}$
• B. $\frac{64}{5}$
• C. $\frac{52}{5}$
• D. $\frac{56}{5}$

Answer 1

The correct option is B $\frac{64}{5}$

$x^2+y^2-8x-4y+16=0$
$(x-4{)}^2+(y-2{)}^2=4\Rightarrow \text{Centre is}(4,2),\text{radius}=2$


$OA=4=OB$
In $\Delta OBC$
$\tan \theta =\frac{4}{2}=2\Rightarrow \sin \theta =\frac{2}{\sqrt{5}}$

In $\Delta BDC$
$\sin \theta =\frac{BD}{2}\Rightarrow BD=\frac{4}{\sqrt{5}}$
$\therefore$ Length of chord of contact, $AB=\frac{8}{\sqrt{5}}$
$(AB{)}^2=\frac{64}{5}$

Alternative:
Length of tangent, $l=4$
and radius, $r=2$
$\Rightarrow$ Length of chord of contact, $AB=\frac{2lr}{\sqrt{l^2+r^2}}=\frac{8}{\sqrt{5}}$
$(AB{)}^2=\frac{64}{5}$