I am not able to figure out this question
What is the locus of the centre of a circle which touches a given line and passes through a given point, not lying on the given line?
I think it's a parabola but I am not able to prove it mathematically
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One of the standard characterizations of a parabola is that it's the set of points equidistant from a specified line --- the directrix --- and a specified point --- the focus. This seems like another way of saying the same thing. – Michael Hardy Apr 04 '14 at 16:32
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Without loss of generality, we may assume that the line is the $x$-axis, and that the point is the point $(0,a)$, where $a$ is positive.
A circle with centre $(p,q)$ that touches the $x$-axis must have radius $|q|$. So it has equation $$(x-p)^2+(y-q)^2=q^2.$$ The circle goes through $(0,a)$. It follows that $$(0-p)^2+(a-q)^2=q^2.$$ Simplify. We get $$p^2+a^2-2aq=0.$$ We can rewrite this as $$q=\frac{1}{2a}p^2 +\frac{a}{2},$$ indeed the equation of a parabola.
André Nicolas
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Assume the line is y-axis and the point is $(a ,0)$.the centre of circle is $(h,k)$ then the equation of circle:
$$(x-h)^2 +(y-k)^2 =r^2 \qquad (1)$$
where $r=$radius of circle but $r=k$ putting this value in equation $(1)$ we get
$$(x-h)^2 +(y-k)^2 =k^2 \qquad (2)$$
but the circle is passing through the point $(a,0)$ therefore putting $x=a, y=0$ in equation $(2)$ we get
$$(a-h)^2 +(0-k)^2 =k^2 $$
simplify:
$$a^2 +h^2 =2ah \qquad (3)$$
equation $(3)$ is the locus of centre of circle touching one axis and passing through a point not lying on the given line.
Eric Stucky
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