Question

The graph of curve
$x^2+y^2-8x-8y+32=0$ falls wholly in the

• A. first quadrant
• B. second quadrant
• C. third quadrant
• D. none of these

Answer 1

The correct option is A first quadrant

$x^2+y^2-2xy-8x-8y+32=0$

$\Rightarrow x^2+y^2-2xy=8x+8y-32$

$\Rightarrow {(x-y)}^2=8(x+y-4)$

is a parabola whose axis is $x-y=0$ and the tangent at the vertex is $x+y-4=0$

Also, when $y=0$, we have

$x^2-8x+32=0$ which gives no real values of $x$

when$x=0$, we have $y^2-8y+32=0$ which gives no real values of $y$.

So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.