MCQ
The circle ${x^2} + {y^2} - 8x + 4y + 4 = 0$ touches
- A$x$-axis only
- ✓$y$- axis only
- CBoth $x$ and $y$- axis
- DDoes not touch any axis
We know that the standard equation of the circle with centre $(h, k)$ is ${(x - h)^2} + {(y - k)^2} = {r^2}.$
Comparing the given equation with the standard equation, we get centre $\equiv $ $(4,\, - 2)$ and radius = $4$.
Since co-ordinates of the centre of the circle are $(4,\, - 2)$, therefore the given circle touches $y$-axis only.
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$\left( {x + 2{y^3}} \right)\frac{{dy}}{{dx}} - y = 0$ is
$2 x-y+2 z=2$
$x-2 y+\lambda z=-4$
$x+\lambda y+z=4$
has no solution. Then the set $S$