Question

The common chord of the circles $x^2+y^2-4x-4y=0$ and $x^2+y^2-16=0$ subtends at the origin an angle equal to

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is D ${90}^0$

We kow that common chord of two circles $S_1$ and $S_2$ is given by

$S_1-S_2=0$

So, common chord of given circles,

$\to (x^2+y^2-4x-4y)-(x^2+y^2-16)=0$

$\to x+y=4$

Let's find the point of intersection of circles by substituting $x=4-y$

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

$\to 2y^2-8y=0$

$\to y=4,0$

$\to x=0,4$

So, intersection points are $(4,0)$ and $(0,4).$

Since, intersection points of circles lie on the x-axis and y-axis,

So, the angle subtended by the chord at origin is ${90}^0.$

Hence, (D) is the correct answer.