Finding the radius of a circle in a graph

AI Thread Summary
The discussion revolves around finding the radius of a circle inscribed within the region bounded by the graph of y = -x^2 - 7x + 12 and the x-axis. Participants identify the vertex of the parabola at (-7/2, 97/4) and discuss the relationship between the circle and the parabola, noting that the circle should be tangent to both the parabola and the x-axis. The radius is suggested to be half the y-coordinate of the vertex, but there is confusion regarding fitting the radius into the form (sqrt(p) + q)/r. The conversation highlights the need to determine the intersection points of the circle and parabola to maximize the area of the inscribed circle.
newchie
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Homework Statement



A circle of maximal area is inscribed in the region bounded by the graph of y = -x^2-7x+12 and the x axis. The radius of this circle is of the form (sqrt(p) + q)/r where, p, q and r are integers and are relatively prime.What is p+q+r

Homework Equations


Vertex form a(x-h)^2+k i believe

The Attempt at a Solution


So i found the vertex, then i assumed that is one point on the circle, and the other is at (same x,0) then shifted along that but i don't seem to get it of the form sqrtp+q all over r
 
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I'd suggest trying to use the distance formula for the radius. Not too sure about this but wouldn't the centre of the circle be at the (h,(k/2))? Would you happen to have the answer (from the solution manual or something)?
 
yes it would be h,k/2
But you get the vertex to be (-97/2,-7/2)
So the radius would just be half the y coordinate, but it doesent really fit the condition of the sqrtp and such part. I think the maximization case i picked is wrong
 
newchie,
You are assuming that the circle and parabola coincide at the vertex of the parabola. I'm not sure that this is true, although it might be. If the curvature of the parabola at the vertex is greater than the curvature of the circle, the circle won't be inside the parabola, hence won't be inscribed in the parabola.

newchie said:
But you get the vertex to be (-97/2,-7/2)
So the radius would just be half the y coordinate, but it doesent really fit the condition of the sqrtp and such part. I think the maximization case i picked is wrong
No, this isn't where the vertex is - it's at (-7/2, + 97/4).
 
Yes my error,

Do you have any ideas on how to maximize this I really am stumped o.op
 
What class are you taking? Is it a calculus class or one that comes before calculus? The type of class you are in will determine the approaches that are available.
 
Mark44 said:
What class are you taking? Is it a calculus class or one that comes before calculus? The type of class you are in will determine the approaches that are available.

Gr 10, this is a challenge problem. However I know a lot of advanced material, but not calculus, so that won't be helpful :/
 
newchie said:

Homework Statement



A circle of maximal area is inscribed in the region bounded by the graph of y = -x^2-7x+12 and the x axis. The radius of this circle is of the form (sqrt(p) + q)/r where, p, q and r are integers and are relatively prime.What is p+q+r

Homework Equations


Vertex form a(x-h)^2+k i believe

The Attempt at a Solution


So i found the vertex, then i assumed that is one point on the circle, and the other is at (same x,0) then shifted along that but i don't seem to get it of the form sqrtp+q all over r
It looks like you have established the vertex of the parabola as being at (-7/2, 97/4).

It makes sense that the circle with maximal area will be tangent to the parabola at two points, both with the same y coordinate. --- or possibly tangent at only one point if that's the vertex. The maximal circle should also be tangent to the x-axis at x = -7/2, the same x value as the vertex.

Let A be the radius of the circle.

You can assume the center of the circle has the same x coordinate as the vertex of the parabola, namely, -7/2 . Then the center of the circle is at (-7/2, A).

What is the equation for such a circle?

Determine where the circle & parabola intersect.
 
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