Identifying the conic given some conditions.

Question

So I have to identify the conic which represents the centre of the circle which touches another circle externally, and also touches the x axis.

Here's a link to the exact question with the equation of the given circle:

https://s24.postimg.org/6yrcii4r9/Screenshot_2016-04-05-00-55-09~01.png

Now the answer as told by many is that its a Parabola.

My argument is that given the symmetric nature of the conic around both sides of the circle, It should be a hyperbola.. It seems I am wrong somewhere, Please help.

Answer 1

The fixed circle has centre $(4,4)$ and radius $6$.

The locus of the centre of the circle is actually two part-parabolas.

When the circle lies above the $x$ axis, the focus of the parabola is at the centre of the circle $(4,4)$ and the directrix is the line $y=-6$

This is because the distance from the centre of the variable circle to the point $(4,4)$ is its radius plus 6, the radius of the fixed circle, and this is equal to the distance from the centre of the variable circle to the line $y=-6$, thus satisfying the geometric definition of a parabola.

Likewise, when the centre of the variable circle is below the $x$ axis, the focus is again at $(4,4)$ and the directrix is the line $y=6$

Answer 2

The answer is (d) : a parabola made of the union of 4 parabolic arcs belonging to parabolas $P_1$ and $P_2$ with a common focus $O(4,4)$ and resp. directrices with equation $y=-6$ for the arcs of parabola above the $x$ axis and $y=6$ for the arcs of parabola below the $x$ axis.

Recall about the focal def. of a parabola: it is the locus of points at equal distance from a fixed point (its focus F) and a fixed straight line (its directrix D).

The equation of circle $(C_0)$ can be put into the form

$$(x-4)^2+(y-4)^2=6^2$$

showing that its center is $C(4,4)$ and its radius is $r_0=6$.

Let us understand what happens for the upper parabola (have a look at the figure): we are looking for centers of circles $C$ that share the common property of being externally tangent to $(C_0)$ and to the $x$ axis in the upper half plane: this gives a constraint $CO=CH$ (with $H$ the orthogonal projection of $C$ onto the straight line with equation $y=-6$) that is always fulfilled, whatever the circle with the above property .

Here is the Matlab program that has given the figure:

clear all;close all;hold on; axis equal
set(gcf,'color','w')
t=0:0.01:2*pi;
cer=@(a,b,r)(plot([a+r*cos(t),NaN,a+0.05*[-1,1,1,-1,-1]],...
[b+r*sin(t),NaN,b+0.05*[-1,-1,1,1,-1]],'linesmoothing','on'));
cer(4,4,6);
for r=0.25:0.25:5
    cer(4+sqrt(20*r+20),r,r)
end
for r=0.25:0.25:4
    cer(4+sqrt(4*r+20),-r,r)
end
plot([-4,18],[6,6])
plot([-4,18],[0,0])
plot([-4,18],[-6,-6])
x=-3:0.01:11;
plot(x,-0.25*(x-4).^2+5,'linesmoothing','on');
plot(x,0.05*(x-4).^2-1,'linesmoothing','on')
a=15.;b=5.;
plot([4,a,a],[4,b,-6],'linesmoothing','on');
set(gca,'defaulttextfontsize',16);
text(a,5.2,'C');text(3.2,4.1,'0');text(a+0.1,-6+0.6,'H')