Question

The length of the common chord of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+6x=0 and ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+3y=0 is
A. $\dfrac{6}{{\sqrt 5 }}$
B. $\dfrac{{\sqrt 6 }}{{10}}$
C. $\dfrac{6}{{10}}$
D. $\dfrac{3}{{\sqrt 5 }}$

Answer 1

Hint: To solve this question we use the basic theory related to the topic common chord between two circles. As we know if we have two circles ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+6x=0 and${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+3y=0. then equation of common chord of the circles can be written as ${{\text{S}}_{\text{1}}}$- ${{\text{S}}_{\text{2}}}$=$0$. And then after using geometry we simply calculate the length of the common chord of the circle.

Complete step-by-step answer:
Let, ${{\text{S}}_{\text{1}}}$: ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+6x=0
${{\text{S}}_{\text{2}}}$: ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+3y=0

As we know,
Equation of common chord is: -
${{\text{S}}_{\text{1}}}$- ${{\text{S}}_{\text{2}}}$=$0$
$\Rightarrow$ ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+6x-(${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+3y)=0
$\Rightarrow$ 6x-3y=0
$\Rightarrow$ 2x=y
$\Rightarrow$ 2x-y=0
Now, we have an equation of the common chord of given circles.
Perpendicular distance of 2x-y= 0 from (-3, 0).
Now, for simplification, we consider a single circle with a common chord of length AC.
Let, the first circle having radius OA and common chord is AC.

Here, OB=$\dfrac{{2 \times ( - 3) - 0}}{{\sqrt {4 + 1} }}$=$\dfrac{{ - 6}}{{\sqrt 5 }}$
And OA=r=3 (given)
Now use Pythagoras for AB.
AB=$\sqrt {{\text{O}}{{\text{A}}^{\text{2}}}{\text{ - O}}{{\text{B}}^{\text{2}}}}$
= $\sqrt {{3^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{ - 6}}{{\sqrt 5 }}} \right)}^{\text{2}}}}$
= $\sqrt {9 - \dfrac{{36}}{5}}$
= $\sqrt {\dfrac{{45 - 36}}{5}}$
= $\dfrac{3}{{\sqrt 5 }}$
Thus, The length of the common chord of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+6x=0 and ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$+3y=0 is$\dfrac{3}{{\sqrt 5 }}$.
Therefore, option (D) is the correct answer.

Note- In this question we have to used Pythagoras theorem which states that, in a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides (i.e. base and height of the given right-angled triangle) “.