For a National Board Exam Review:
Point (3,4) is the center of the circle tangent to the x-1 axis. What is the point of tangency?
Answer is (3,0)
I usually would provide an attempt but I do not understand the problem? How can the center of a circle be tangent to a line when only curves can be tangent to something?
Is the problem set wrong? How do I visualize this?
$\endgroup$ 34 Answers
$\begingroup$I assume that the question means the $x$-axis.
In geometry, a line is tangent to a circle if the line intersects the circle in exactly one point. This concept is generalized in calculus, but this question seems to use the simple geometry concept. I suppose you could also say that the circle is tangent to the line.
You can see in this diagram that the point of tangency between the circle with center $(3,4)$ and the $x$-axis is indeed the point $(3,0)$. The question is testing if you can visualize this.
$\endgroup$ $\begingroup$The circle is tangent to the $x$-axis (although I would prefer to say that the $x$-axis is tangent to the circle). The centre of the circle is (3,4).
You can visualise this as a circle rolling along the $x$-axis.
$\endgroup$ $\begingroup$The point of tangency will be the closest point to the circle on the X Axis. As a rule of thumb, the shortest distance between a point (3,4) and a line (x=0) will be perpendicular to the line and run through the point. In this case, graph a vertical line that goes through your point (3,4) and you'll see that it hits the X Axis at (3,0). That's your point of tangency.
$\endgroup$ $\begingroup$Since, the circle, having center at $(3, 4)$, touches the x-axis hence the radius of the circle $$=\text{normal distance from the x-axis}=4$$
Now, the equation of the circle having center $(3, 4)$ & a radius $4$ is given as $$(x-3)^2+(y-4)=4^2$$ $$x^2-6x+9+y^2-8y+16=16$$ $$x^2+y^2-6x-8y+9=0$$ Now, at the point of tangency with x-axis, $y=0$ hence, setting $y=0$ in the equation of the circle we get $$x^2-6x+9=0$$ $$(x-3)^2=0\iff x=3, 3$$ Hence the circle will touch the x-axis at $(3, 0)$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Point of tangencey:}\ \color{blue}{(3, 0)}}}$$
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