Question

𝑥2 + 𝑦2 − 8𝑥 + 2𝑦 − 32 = 0?

Answer 1

Since the question is incomplete, we find the type of conic section and its parameter.

Solution:

$𝑥^2 + 𝑦^2 − 8𝑥 +2𝑦 − 32 = 0 \\ \Rightarrow x^2-8x+y^2+2y=32 \\ \Rightarrow (x-4)^2-4^2+(y+1)^2-1^2=32 \\ \Rightarrow (x-4)^2+(y+1)^2=32+17 \\ \Rightarrow (x-4)^2+(y+1)^2=49 \\\Rightarrow (x-4)^2+(y+1)^2=7^2$

It is a circle. On comparing with $(x-h)^2+(y-k)^2=r^2$, we get,

Centre$=(h,k)=(4,-1)$ and radius$=r=7$ units.