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Q.
Find the centre and the radius of the circle $x^{2}+y^{2}+8x+10y-8=0$.
Conic Sections
Solution:
The given equation is $\left(x^{2} + 8x\right) + \left(y^{2} + 10y\right) = 8$
Now, completing the squares within the parenthesis, we get
$\left( x^{2} + 8 x + 16\right) + \left( y^{2}+ 10y + 25\right) = 8 + 16 + 25$
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 $7$.