Question

If the equation of circle is $x^2+y^2+8 x+10 y-8=0$, then
I. The centre is $(-4,-5)$.
II. The radius is 7.

• A. Both I and II are true
• B. Only I is true
• C. Only II is true
• D. Both I and II are false

Answer 1

Answer: (A)

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