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 ${\left(AB\right)}^2$ 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

$\frac{64}{5}$

Explanation for the correct option:

Finding the value of ${\left(AB\right)}^2$:

Given the equation of a circle is $x^2+y^2-8x-4y+16=0$ which touches it at the points $A$ and $B$.

By comparing the equation of a circle with the general equation of $x^2+y^2+2g+2f+c=0$ we get,

$\Rightarrow g=-4;f=-2\text{and}c=16$

The radius of a circle is calculated using $R=\sqrt{g^2+f^2-c}$.

$\begin{array}{rcl}R& =& \sqrt{{\left(-4\right)}^2+{\left(-2\right)}^2-16}\\ & =& \sqrt{16+4-16}\\ & =& \sqrt{4}\\ R& =& 2\end{array}$

And $L$ is calculated as $L=\sqrt{S_1}$.

$\begin{array}{rcl}L& =& \sqrt{S_1}\\ & =& \sqrt{16}\\ L& =& 4\end{array}$

The length of the chord contact $AB$ is calculated as,

$\Rightarrow AB=\frac{2LR}{\sqrt{L^2+R^2}}$

Substitute the value of $L$ as $4$ and $R$ as $2$ in the length of the chord $AB$.

$\begin{array}{rcl}AB& =& \frac{2\left(4\right)\left(2\right)}{\sqrt{{\left(4\right)}^2+{\left(2\right)}^2}}\\ & =& \frac{2\times 8}{\sqrt{4+16}}\\ & =& \frac{2\times 8}{\sqrt{20}}\\ & =& \frac{2\times 8}{2\sqrt{5}}\\ AB& =& \frac{8}{\sqrt{5}}\end{array}$

By squaring both sides we obtain the value of ${\left(AB\right)}^2$.

$\begin{array}{rcl}{\left(AB\right)}^2& =& {\left(\frac{8}{\sqrt{5}}\right)}^2\\ & =& \frac{64}{5}\end{array}$

Therefore, the value of ${\left(AB\right)}^2$ is $\frac{64}{5}$.

Hence, option (B) is the correct answer.