Question

Find the center and radius of the circle $x^2 + y^2 - 8x + 10y - 12 = 0$

Answer 1

_Given that _

The equation of the circle $x^2 + y^2 - 8x + 10y - 12 = 0$

Then we know that The standard equation of the circle is $(x - h)^2 + (y -k)^2 = r^2$

$x^2 + y^2 - 8x + 10y - 12 = 0$

$⇒x^2+4^2-8x+y^2+5^2+10y-41-12=0$

$⇒x^2+4^2-8x+y^2+5^2+10y-53=0$

$⇒(x-4)^2+(y+5)^2=(√53)^2$

Now comparing the above equation with the standard equation we get

The circle of the equation is $(4,-5)$ and the radius is $√53$ (Ans.)