Question

The common chord of the circles $x^2+y^2-4x-4y=0$ and $2x^2+2y^2=32$ subtends at the origin an angle equal to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

The given circles are $x^2+y^2-4x-4y=0$ ..... $\left(i\right)$ and $2x^2+2y^2=32$ i.e. $x^2+y^2=16$ ....... $\left(ii\right)$

To find the common chord we need to find the intersection points of the circles

So, let's subtract $\left(ii\right)$ from $\left(i\right)$, we get

$-4x-4y=-16$

i.e. $x+y=4$ ........ $\left(iii\right)$

Substitute $x=y-4$ in $\left(ii\right)$, we get

$(y-4{)}^2+y^2=16$

$⟹y^2-8y+16+y^2=16$

$⟹y^2-4y=0$

$⟹y(y-4)=0$

$⟹y=0,4$

At $y=0,x=4$ and at $y=4,x=0$

So, the points of intersection of these two circles is $(4,0)$ and $(0,4)$

Thus, the chord drawn from $(0,4)$ to $(4,0)$

Thus, the angle subtended at the origin by this chord is a right angle.

Hence, the chord subtends an angle of $\frac{\pi }{2}$ at the origin.