Parabola : To prove that Locus of centre is a parabola.:?

Take the given circle as x^2 + y^2= r^2 and line as y= mx + c where distance of this line from origin is greater than r. Let the centre of the variable circle ve (h,k).

So,

Since it touches the fixed circle, so if draw a rough fig. you will see that distance of h,k from origin is radius of fixed circle + radius of variable circle i.e.

sqrt((h-0)^2+(k-0)^2)= r+ r' whre r' is rad of variable circle.

now, r' is also equal to dist of h,k from the line.

r'=| mh-k+c|/sqrt(m^2+1)

subst this in above qn and square, you will get the eqn of a parabola.

At first thought I expected the locus to be a hyperbola. The description seems to describe a curve that approaches the diagonals as the radius of the sequence of circles increases without bound.

A parabola does not approach a straight line.

It is definitely not a parabola but I have not proved it is a hyperbola.

a million) I deleted my grossly incorrect attempt to unravel this question and could attempt it again later. it truly is actual that I mistook a and b as constants which became a gross blunders. could be that Falzoon is nice in concluding it to be an ellipse which could be a circle as a particular case of an equilateral triangle, yet i want to workout elementary the perfect answer. Edit: My partial answer of one million) is as below. Lengths of aspects of triangle ABC are consistent. enable A = (m, 0), B = (0, n) and C = (x, y) => x^2 + (y - n)^2 = a^2 ... ( a million ) (x - m)^2 + y^2 = b^2 ... ( 2 ) and m^2 + n^2 = c^2 ... ( 3 ) eliminating the variables m and n from the three equations could provide the equation of the locus. i could no longer attempt this very last step. Edit: I actually have tried to get rid of m and n as below: From eqn. ( a million ) and ( 2 ), m = x ± ?(b^2 - y^2) and n = y ± ?(a^2 - x^2) Plugging in eqn. ( 3 ), [x ± ?(b^2 - y^2)]^2 + [y ± ?(a^2 - x^2)]^2 = c^2 => x^2 + b^2 - y^2 ± 2x?(b^2 - y^2) + y^2 + a^2 - x^2 ± 2y?(a^2 - x^2) = c^2 => 2x?(b^2 - y^2) + 2y?(a^2 - x^2) = ± (c^2 - a^2 - b^2) => 4x^2 (b^2 - y^2) + 4y^2 (a^2 - x^2) + 8xy ?[(b^2 - y^2)(a^2 - x^2)] = (c^2 - a^2 - b^2)^2 => b^2x^2 + a^2y^2 - 2x^2y^2 + 2xy?[(b^2 - y^2)(a^2 - x^2)] = ok (a consistent). Edit: Falzoon's prediction of the locus being ellipse could nicely be inferred as below. we may be able to imagine of the vertex C of the triangle ABC to be very close to to the line AB so as that C is almost on the line AB. thus, the locus of C is an ellipse could nicely be proved actually. From this, it may nicely be inferred that the locus is elliptical or almost so. Edit: i tried your advice for triangle, sq. and hexagon and curiously discovered the integer solutions that were an same because the variety of aspects, 3 for triangle, 4 for sq. and six for hexagon. exciting!! I shall attempt this example sometime as we communicate when I get sufficient time. Edit: I checked utilizing Wolfram Alpha for n = 3 to 11 that the answer equals the variety of aspects of the polygon, yet could no longer go with a time-honored information for any fee of n.

even tho im in precal, i still dont know how to do this.

but this is wat i think:

imagine circle #1

in the center, theres a line going across

now on the center is a point of circle #2

circle #2 has two points intersecting circle #1 and a point in the center of circle #1

if you can draw this out you can see how theres a parabola. idk how to prove the locus tho. Sorry