Question

A circle of area $16\pi$ square units has two of its diameters along the lines $x+y=5$ and $x-y=1$. Find the equation of the circle.

Answer 1

I) circle area $=16\pi$

$\pi r^2=16\pi$

$r^2=16$

$r=4units$

Diameter along lines $x+y=5$

$x-y=1$

Both lines are perpendicular to each other and center is the point

of intersection of lines

$y=-x+5$

So $x-y=1$

$x+x-5=1$

$2x=6$

$x=3$

So, $y=2$

Equation of circle $\Rightarrow (x-3{)}^2+(y-2{)}^2=16$