I have the following question
As you can see on the image, the equation of the circle is $(x+2)^2 + (y+4)^2 = 4$
What I don't understand is how can I check the II. and III. propositions. The answer sheet says that only the III proposition is true, graphing shows otherwise
How can I check the II. and III. options through calculation ? And not graphing? And why isn't the circle tangent to the x-axis?
$\endgroup$ 33 Answers
$\begingroup$The equation of a circle is $(x−h)^2+(y−k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. The correct graph is a circle centered at $(-2,-4)$.
If the circle $(x+2)^2+(y+4)^2=4$ is tangent to the x-axis, then it must intersect or touch the x-axis. As $y=0$ on the x-axis, we need to satisfy $(x+2)^2+(y+4)^2=4$ when $y=0$. Substituting $y=0$, we obtain $(x+2)^2=0$ or $x=-2$, which is the point $(-2,0)$.
Then, since our circle is centered at $(-2,-4)$, it must have a radius of $4$ in order to be big enough to reach the x-axis. However, our circle only has radius $2$, so it cannot reach the x-axis. So, II is false.
III is true because the circle does touch the y-axis. When we substitute $x=0$, we obtain $(y+4)^2=0$ or $y=-4$, which is the point $(-4,0)$. As a circle is centered at $(-2,-4)$ and has a radius of $2$, it is big enough to reach the y-axis at the point $(0,-4)$. At this point, the circle will be tangent to the y-axis.
$\endgroup$ $\begingroup$If the circle is tangent to the x-axes, the the $y$ coordinate must be zero, this is impossible, then $$(x+2)^2+4^2\neq 0$$. The circle is tangent to the y axes is true, since we have $x=0$ then we get $$4+(y+2)^2=4$$ so $y=-2$
$\endgroup$ $\begingroup$The center of the circle is (-2, -4). So, the distance from the center to the $y$-axis and 2 and to the $x$-axis is 4.
Since the radius of the circle is 2, it is only tangential to the $y$-axis because the distance is the same as the radius.
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