Question

The length of intercept cut off from the line y = mx +c by the circle ${\text{x}}^2+{\text{y}}^2=a^2$ is
A.$\sqrt{{\text{a}}^2\left(1+m^2\right)-c^2}$
B. ${\displaystyle \frac{\sqrt{{\text{a}}^2\left(1+m^2\right)-c^2}}{\sqrt{1+m^2}}}$
C. ${\displaystyle \frac{2\sqrt{{\text{a}}^2\left(1+m^2\right)-c^2}}{\sqrt{1+m^2}}}$
D. ${\text{a}}^2\left(1+m^2\right)-c^2$

Answer 1

Hint: In this question, first we will make the diagram with the centre of the circle as (0,0) and radius as ‘a’. After this we will draw a perpendicular from centre to chord form by the given line. Then determine the length of the perpendicular and use it to calculate the length of chord.

Complete step-by-step answer:
The diagram for the question is:

OM is the perpendicular drawn from centre o. It divides AB into two parts such that AM = BM.
OA is the radius = a

We know that length of perpendicular drawn from point(${\text{x}}_1,{\text{y}}_1$) is given by:
d = ${\displaystyle \frac{\text{|a}{\text{x}}_1+\text{b}{\text{y}}_1+c|}{\sqrt{{\text{a}}^2+{\text{b}}^2}}}$
Therefore, we can say that:
OM = ${\displaystyle \frac{c}{\sqrt{1+m^2}}}$
Using Pythagoras theorem, we can write:
$\text{A}{\text{M}}^2=\text{O}{\text{A}}^2-\text{O}{\text{M}}^2$
Putting the values in above equation, we get:
$\text{A}{\text{M}}^2={\text{a}}^2-{\left({\displaystyle \frac{c}{\sqrt{1+m^2}}}\right)}^2$
$\Rightarrow$ $\text{AM}=\sqrt{{\text{a}}^2-{\left({\displaystyle \frac{c}{\sqrt{1+m^2}}}\right)}^2}=\sqrt{{\text{a}}^2-{\displaystyle \frac{c^2}{{(1+m^2)}^{}}}}=\sqrt{{\displaystyle \frac{{\text{a}}^2(1+m^2)-c^2}{{(1+m^2)}^{}}}}$
Therefore, length chord AB = $2\sqrt{{\displaystyle \frac{{\text{a}}^2(1+m^2)-c^2}{{(1+m^2)}^{}}}}$
So, option C is correct.

Note- In the question involving finding the length of the chord, you should remember the formula for finding the length of perpendicular drawn from a point(${\text{x}}_1,{\text{y}}_1$) which is given by:
d = ${\displaystyle \frac{\text{|a}{\text{x}}_1+\text{b}{\text{y}}_1+c|}{\sqrt{{\text{a}}^2+{\text{b}}^2}}}$ . You should know that the perpendicular drawn from centre on the chord divides the chord into two equal parts.