Question

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

• A. $(5,4),-7$
• B. $(-4,-5),7$
• C. $(4,-5),7$
• D. $(5,-4),7$

Answer 1

Answer: (B)

The given equation is $\left(x^2+8x\right)+\left(y^2+10y\right)=8$
Now, completing the squares within the parenthesis, we get
$\left(x^2+8x+16\right)+\left(y^2+10y+25\right)=8+16+25$
i.e., ${\left(x+4\right)}^2+{\left(y+5\right)}^2=49$ i.e. ${\left(x-\left(-4\right)\right)}^2+{\left(y-\left(-5\right)\right)}^2=7^2$
Therefore, the given circle has centre at $\left(-4,-5\right)$ and radius $7$.