Question

The angle subtended by the common chord of the circles ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 4x - 4y = 0}}$ and ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 16}}$ at the origin is-
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{\pi }{2}$

Answer 1

Hint: To solve this question we use the basic theory related to the topic of common chord between two circles. As we know if we have two circles ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 4x - 4y = 0}}$ and ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 16}}$. then equation of common chord of the circles can be written as ${{\text{S}}_{\text{1}}}$- ${{\text{S}}_{\text{2}}}$=$0$. So, by using this we get our desired result.

Complete step-by-step answer:
As mentioned in question, we have two circles.
Let, ${{\text{S}}_{\text{1}}}$: ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 4x - 4y = 0}}$
       ${{\text{S}}_{\text{2}}}$: ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 16}}$

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}}}{\text{ - 4x - 4y}}$- (${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - 16}}$)=0
$ \Rightarrow $−4x−4y+16=0
$ \Rightarrow $x+y=4
which is a line equally inclined to the axes.
∴ Angle subtended by the common chord at origin is $\dfrac{\pi }{2}$.
Thus, option (D) is correct.


Note- Common chord of two intersecting circles is the chord which is common to both the circles. We can also say; the common chord of two intersecting circles is the line segment joining points of intersection of two circles as shown in the above figure.