Question

If the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ bisects the circumference of the circle ${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$, then the length of the common chord of the circles is
(a) $2\sqrt{g_{1}^{2}+f_{1}^{2}-{{c}_{1}}}$
(b) $\sqrt{g_{1}^{2}+f_{1}^{2}-{{c}_{1}}}$
(c) $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
(d) $2\sqrt{{{g}^{2}}+{{f}^{2}}-c}$

Answer 1

Hint: To solve the above question, we have to know the concept of the length of common chords of circles. As we can see ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ bisect the circumference of the circle with equation
${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$ Then the length of common chord will be equal to the diameter of circle equation ${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$ .

Complete step-by-step solution:
We can see that when two circles intersect, then we can connect the two intersection points and made a common chord.
Since we have the circle with equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ bisect the circumference of the circle with equation ${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$ then the length of common chord will be equal to
Diameter of circle of common chord.

Now we can see how to find diameter of a circle
If we consider the equation the circle equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+d=0$ whose center is $\left( -g,-f \right)$
So, ${{x}^{2}}+{{y}^{2}}+2gx+2fy+d=0$
Now make it square form
$\Rightarrow \left( {{x}^{2}}+2gx+{{g}^{2}} \right)+\left( {{y}^{2}}+2fy+{{f}^{2}} \right)={{g}^{2}}+{{f}^{2}}-d$
Now make it simplify and we will get the equation as,
${{\left( x+g \right)}^{2}}+{{\left( y+f \right)}^{2}}={{g}^{2}}+{{f}^{2}}-d$
Now we make it in circle form like ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$
So, we will get the equation as,
${{\left\{ x-\left( -g \right) \right\}}^{2}}+{{\left\{ y-\left( -f \right) \right\}}^{2}}={{\left( \sqrt{{{g}^{2}}+{{f}^{2}}-d} \right)}^{2}}$
So, here we can see the center is $\left( -g,-f \right)$ and diameter is = $2\times \sqrt{{{g}^{2}}+{{f}^{2}}-d}\ldots \left( 1 \right)$
Now if we consider our equation ${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0$
Here if we comparing with $\left( 1 \right)$ We will get,
$g={{g}_{1,}}f={{f}_{1}}$ And $d={{c}_{1}}$
So, we will get the length of common chord as,
$2\times \sqrt{g_{1}^{2}+f_{1}^{2}-{{c}_{1}}}$
Hence, the correct option is (a)$2\sqrt{g_{1}^{2}+f_{1}^{2}-{{c}_{1}}}$

Note: Here students must take care of the concept of length of common chords of circles. Student sometimes did a mistake because they are only considering the formula of radii $\sqrt{g_{1}^{2}+f_{1}^{2}-{{c}_{1}}}$
Forget to multiply with $2$ as $Diameter=2\times Radii$. So, the Student has to take care of it.